# On the generalised sum of squared logarithms inequality

## Abstract

Assume $$n\geq2$$. Consider the elementary symmetric polynomials $$e_{k}(y_{1},y_{2},\ldots, y_{n})$$ and denote by $$E_{0},E_{1},\ldots,E_{n-1}$$ the elementary symmetric polynomials in reverse order $$E_{k}(y_{1},y_{2},\ldots,y_{n}):=e_{n-k}(y_{1},y_{2},\ldots,y_{n})= \sum_{i_{1}<\cdots<i_{n-k}} y_{i_{1}}y_{i_{2}}\cdots y_{i_{n-k}}$$, $$k\in\{0,1,\ldots,n-1 \}$$. Let, moreover, S be a nonempty subset of $$\{0,1,\ldots,n-1\}$$. We investigate necessary and sufficient conditions on the function $$f:I\to\mathbb{R}$$, where $$I\subset\mathbb{R}$$ is an interval, such that the inequality $$f(a_{1})+f(a_{2})+\cdots+f(a_{n})\leq f(b_{1})+f(b_{2})+\cdots+f(b_{n})$$ () holds for all $$a=(a_{1},a_{2},\ldots,a_{n})\in I^{n}$$ and $$b=(b_{1},b_{2},\ldots,b_{n})\in I^{n}$$ satisfying $$E_{k}(a)< E_{k}(b)$$ for $$k\in S$$ and $$E_{k}(a)=E_{k}(b)$$ for $$k\in\{0,1,\ldots,n-1\}\setminus S$$. As a corollary, we obtain our inequality () if $$2\leq n\leq4$$, $$f(x)=\log^{2}x$$ and $$S=\{1,\ldots,n-1\}$$, which is the sum of squared logarithms inequality previously known for $$2\le n\le3$$.

## 1 Introduction - the sum of squared logarithms inequality

In a previous contribution  the sum of squared logarithms inequality has been introduced and proved for the particular cases $$n=2,3$$. For $$n=3$$ it reads: let $$a_{1},a_{2},a_{3},b_{1},b_{2},b_{3}>0$$ be given positive numbers such that

\begin{aligned}& a_{1}+a_{2}+a_{3} \le b_{1}+b_{2}+b_{3}, \\& a_{1} a_{2}+a_{1} a_{3}+ a_{2} a_{3} \le b_{1} b_{2}+b_{1} b_{3}+ b_{2} b_{3}, \\& a_{1} a_{2} a_{3} =b_{1} b_{2} b_{3}. \end{aligned}

Then

$$\log^{2} a_{1}+\log^{2} a_{2}+ \log^{2} a_{3} \le \log^{2} b_{1}+ \log^{2} b_{2}+\log ^{2} b_{3} .$$

The general form of this inequality can be conjectured as follows.

### Definition 1.1

The standard elementary symmetric polynomials $$e_{1},\ldots ,e_{n-1}, e_{n}$$ are

$$e_{k}(y_{1},\ldots,y_{n})=\sum _{1\le j_{1}< j_{2}<\cdots<j_{k}\le n} y_{j_{1}}\cdot y_{j_{2}}\cdot\, \cdots\,\cdot y_{j_{k}} ,\quad k\in\{1,2,\ldots,n\} ;$$
(1.1)

note that $$e_{n}=y_{1}\cdot y_{2}\cdot\,\cdots\,\cdot y_{n}$$.

### Conjecture 1.2

(Sum of squared logarithms inequality)

Let $$a_{1},a_{2},\ldots,a_{n}$$, $$b_{1},b_{2},\ldots,b_{n}$$ be given positive numbers. Then the condition

$$e_{k}(a_{1},\ldots, a_{n}) \le e_{k}(b_{1},\ldots, b_{n}),\quad k\in\{ 1,2, \ldots,n-1\},\qquad e_{n}(a_{1},\ldots, a_{n}) = e_{n}(b_{1},\ldots, b_{n})$$

implies that

$$\sum_{i=1}^{n} \log^{2} a_{i} \le \sum_{i=1}^{n} \log^{2} b_{i} .$$

### Remark 1.3

Note that the conclusions of Conjecture 1.2 are trivial provided we have equality everywhere, i.e.

$$e_{k}(a_{1},\ldots, a_{n}) = e_{k}(b_{1},\ldots, b_{n}), \quad k\in\{1,2,\ldots ,n\} .$$
(1.2)

In this case, the coefficients $$a_{1},\ldots, a_{n}$$, $$b_{1},\ldots, b_{n}$$ are equal up to permutations, which can be seen by looking at the characteristic polynomials of two matrices with eigenvalues $$a_{1},\ldots ,a_{n}$$ and $$b_{1},\ldots,b_{n}$$. From this perspective, having equality just in the last product $$e_{n}$$ and strict inequality else seems to be the most difficult case.

Based on extensive random sampling on $$\mathbb{R}_{+}^{n}$$ for small numbers n it has been conjectured that Conjecture 1.2 might be true for arbitrary $$n\in\mathbb{N}$$. The sum of squared logarithms inequality has immediate important applications in matrix analysis (; see also ) as well as in nonlinear elasticity theory . In matrix analysis it implies that the global minimiser over all rotations to

$$\inf_{Q\in\operatorname{SO}(n)} \bigl\Vert \operatorname{sym}_{*} \operatorname{Log} Q^{T} F\bigr\Vert ^{2}=\bigl\Vert \sqrt{F^{T}F}\bigr\Vert ^{2}$$
(1.3)

at given $$F\in\operatorname{GL}^{+}(n)$$ is realised by the orthogonal factor $$R=\operatorname{polar}(F)$$ (such that $$R^{T} F=\sqrt{F^{T}F}$$). Here, $$\| X\|^{2}:=\sum_{i,j=1}^{n} X_{ij}^{2}$$ denotes the Frobenius matrix norm and $$\operatorname{Log}: \operatorname{GL}(n) \to\mathfrak{gl}(n)=\mathbb{R}^{n\times n}$$ is the multivalued matrix logarithm, i.e. any solution $$Z=\operatorname{Log} X\in\mathbb{C}^{n\times n}$$ of $$\exp(Z)=X$$ and $$\operatorname{sym}_{*}(Z)=\frac{1}{2} ( Z^{*}+Z )$$.

Recently, the case $$n=2$$ was used to verify the polyconvexity condition in nonlinear elasticity [4, 5] for a certain class of isotropic energy functions. For more background information on the sum of squared logarithms inequality we refer the reader to .

In this paper we extend the investigation as to the validity of Conjecture 1.2 by considering arbitrary functions f instead of $$f(x)=\log^{2} x$$. We formulate this more general problem and we are able to extend Conjecture 1.2 to the case $$n=4$$. The same methods should also be useful for proving the statement for $$n=5,6$$. However, the necessary technicalities prevent us from discussing these cases in this paper.

In addition, we present ideas which might be helpful in attacking the fully general case, namely arbitrary f and arbitrary n.

## 2 The generalised inequality

In order to generalise Conjecture 1.2 in the directions hinted at in the introduction, we consider from now on a non-standard definition of the elementary symmetric polynomials. In fact, for $$n\geq2$$ it will be more convenient for us to reverse their numbering and define $$E_{0},E_{1},\ldots,E_{n-1}$$ by

$$E_{k}(y_{1},\ldots, y_{n}):=e_{n-k}( y_{1},\ldots, y_{n})=\sum_{i_{1}< \cdots <i_{n-k}} y_{i_{1}}\cdot y_{i_{2}}\cdot\,\cdots\,\cdot y_{i_{n-k}} , \quad k\in\{0,1,\ldots,{n-1}\} .$$
(2.1)

In particular, now

\begin{aligned} &E_{0}(y_{1},\ldots, y_{n}):=e_{n}( y_{1},\ldots, y_{n})=y_{1} \cdot y_{2}\cdot\, \cdots\,\cdot y_{n} , \\ &E_{n-1}(y_{1},\ldots, y_{n}):=e_{1}( y_{1},\ldots, y_{n})=y_{1}+y_{2}+\cdots +y_{n} . \end{aligned}
(2.2)

Let $$I\subset\mathbb{R}$$ be an open interval and let

$$\Delta_{n}:=\bigl\{ y=(y_{1},y_{2}, \ldots,y_{n})\in I^{n} \mid y_{1}\leq y_{2}\leq \cdots\leq y_{n}\bigr\} .$$
(2.3)

Let S be a nonempty subset of $$\{0,1,\ldots,n-1\}$$ and assume that $$a,b\in\Delta_{n}$$ are such that

$$E_{k}(a)< E_{k}(b)\quad \mbox{for } k\in S \quad \mbox{and}\quad E_{k}(a)=E_{k}(b) \quad \mbox{for } k\in \{0,1,\ldots,n-1\}\setminus S .$$
(2.4)

In this section we investigate necessary and sufficient conditions for a (smooth) function $$f:I\to\mathbb{R}$$, such that the inequality

$$f(a_{1})+f(a_{2})+\cdots+f(a_{n})\leq f(b_{1})+f(b_{2})+\cdots+f(b_{n})$$

holds for all $$a,b\in\Delta_{n}$$ satisfying assumption (2.4).

### Remark 2.1

The formulation of the above problem has a certain monotonicity structure: we assume that ‘$$E(a)< E(b)$$’ and want to prove that ‘$$F(a)< F(b)$$’. Therefore our idea is to consider a curve y connecting the points a and b, such that $$E(y(t))$$ ‘increases’. Then the function $$g(t)=F(y(t))$$ should also increase and therefore $$g'(t)>0$$ must hold. From this we are able to derive necessary and sufficient conditions on the function f.

This approach motivates the following definition.

### Definition 2.2

(b dominates a, $$a\preceq b$$)

Let $$a,b\in\Delta_{n}$$. We will say that b dominates a and denote $$a\preceq b$$ if there exists a piecewise differentiable mapping $$y:[0,1]\to\Delta_{n}$$ (i.e. y is continuous on $$[0,1]$$ and differentiable in all but at most countably many points) such that $$y(0)=a$$, $$y(1)=b$$, $$y_{i}(t)\neq y_{j}(t)$$ for $$i\neq j$$ and all but at most countably many $$t\in[0,1]$$ and the functions

$$A_{k}(t):=E_{k}\bigl(y(t)\bigr) ,\quad k\in\{0,1,\ldots,n-1 \}$$

are nondecreasing on the interval $$[0,1]$$.

If $$a\preceq b$$, then $$E_{k}(a)=A_{k}(0)\leq A_{k}(1)=E_{k}(b)$$, so it follows from Definition 2.2 that a, b satisfy assumption (2.4) with S being the set of all k for which $$A_{k}(t)$$ is not a constant function on $$[0,1]$$.

We are ready to formulate the main results of this section.

### Theorem 2.3

Assume that $$a,b\in\Delta_{n}$$ and let $$a\preceq b$$. Let $$S\subseteq\{0,1,\ldots,n-1\}$$ denote the set of all integers k with $$E_{k}(a)< E_{k}(b)$$. Moreover, assume that $$f\in C^{n}(I)$$ be such that

$$(-1)^{n+k}\bigl(x^{k} f'(x) \bigr)^{(n-1)}\leq0 \quad \textit{for all }x\in I \textit { and all } k\in S .$$
(2.5)

Then the following inequality holds:

$$f(a_{1})+f(a_{2})+\cdots+f(a_{n}) \leq f(b_{1})+f(b_{2})+\cdots+f(b_{n}) .$$
(2.6)

A partially reverse statement is also true.

### Theorem 2.4

Let $$f\in C^{n}(I)$$ be such that the inequality

$$f(a_{1})+f(a_{2})+\cdots+f(a_{n})\leq f(b_{1})+f(b_{2})+\cdots+f(b_{n})$$
(2.7)

holds for all $$a,b\in\Delta_{n}$$ satisfying

$$E_{k}(a)\leq E_{k}(b) \quad \textit{for } k\in S\quad \textit{and}\quad E_{k}(a)=E_{k}(b)\quad \textit{for } k\in\{0,1,\ldots,n-1\}\setminus S$$
(2.8)

for some subset $$S\subseteq\{0,1,\ldots,n-1\}$$. Then f satisfies property (2.5), i.e.

$$(-1)^{n+k}\bigl(x^{k} f'(x) \bigr)^{(n-1)}\leq0\quad \textit{for all }x\in I \textit { and all } k\in S .$$
(2.9)

In this respect, we can formulate another conjecture.

### Conjecture 2.5

Let S be a nonempty subset of $$\{0,1,\ldots,n-1\}$$ and assume that $$a,b\in\Delta_{n}$$ are such that (2.4) is satisfied, i.e.

$$E_{k}(a)< E_{k}(b) \quad \textit{for } k\in S \quad \textit{and} \quad E_{k}(a)=E_{k}(b) \quad \textit{for } k\in \{0,1,\ldots,n-1\}\setminus S .$$

Then there exists a curve y satisfying the conditions from Definition  2.2 and thus $$a\preceq b$$.

### Remark 2.6

In concrete applications of Theorem 2.3 and Theorem 2.4 one would like to know whether condition (2.4) already implies $$a\preceq b$$. This is Conjecture 2.5. Unfortunately, we are able to prove Conjecture 2.5 only for $$2\leq n\leq4$$, $$I=(0,\infty)$$ and $$S\subseteq\{1,2,\ldots,n-1\}$$ (see the next section).

### Example 2.7

It is easy to see that if $$I=(0,\infty)$$ then the function $$f(x)=\log^{2}x$$ satisfies property (2.5) for $$S=\{ 1,2,\ldots,n-1\}$$. Indeed, we proceed by induction on n. For $$n=2$$ and $$k=1$$ the property is immediate. Moreover, for $$k\geq2$$ and $$n\geq3$$ we get

\begin{aligned} (-1)^{n+k}\bigl(x^{k} f'(x) \bigr)^{(n-1)}&=2(-1)^{n+k}\bigl(x^{k-1}\log x \bigr)^{(n-1)} \\ &=2(-1)^{n+k}\bigl((k-1)x^{k-2}\log x\bigr)^{(n-2)}+2(-1)^{n+k} \bigl(x^{k-2}\bigr)^{(n-2)}\leq0 \end{aligned}
(2.10)

by the induction hypothesis, since the second summand vanishes. It remains to check property (2.5) for $$k=1$$, which is also immediate.

Note also that property (2.5) is not true for $$k=0$$. Therefore Theorem 2.3 and Theorem 2.4 for $$f(x)=\log^{2}x$$ attain the following formulation.

### Corollary 2.8

Assume that $$a,b\in\mathbb{R}_{+}^{n}$$ be such that $$a\preceq b$$ and $$a_{1}a_{2}\cdots a_{n}=b_{1}b_{2}\cdots b_{n}$$. Then

$$\log^{2}(a_{1})+\log^{2}(a_{2})+\cdots+ \log^{2}(a_{n})\leq \log^{2}(b_{1})+ \log^{2}(b_{2})+\cdots+\log^{2}(b_{n})$$

and this inequality fails if the constraint $$a_{1}a_{2}\cdots a_{n}=b_{1}b_{2}\cdots b_{n}$$ is replaced by the weaker one $$a_{1}a_{2}\cdots a_{n}\leq b_{1}b_{2}\cdots b_{n}$$.

In order to see that the weaker condition is not sufficient for the inequality to hold, consider the case

$$a = \biggl(\frac{1}{n},\ldots,\frac{1}{n}\biggr) , \qquad b=(1, \ldots,1) .$$

Then $$a \preceq b$$ and $$a_{1}a_{2}\cdots a_{n}\leq b_{1}b_{2}\cdots b_{n}$$, but

$$\log^{2}(a_{1})+\log^{2}(a_{2})+\cdots+ \log^{2}(a_{n}) = n \log^{2}(n) > 0 = \log^{2}(b_{1})+\log^{2}(b_{2})+\cdots+ \log^{2}(b_{n}) .$$

### Remark 2.9

Corollary 2.8 is a weaker statement than Conjecture 1.2 since we assume that $$a\preceq b$$. If Conjecture 2.5 is true, then Conjecture 1.2 follows.

### Example 2.10

The function $$f(x)=x^{p}$$ ($$x>0$$) with $$p\in(0,1)$$ satisfies property (2.5) for the set $$S=\{ 0,1,\ldots,n-1\}$$. Indeed, for each $$n\geq2$$ and $$0\leq k\leq n-1$$, we have

$$(-1)^{n+k}\bigl(x^{k}f'(x)\bigr)^{(n-1)}= (-1)^{n+k}p(k+p-1) (k+p-2)\cdots\bigl(k+p-(n-1)\bigr)x^{k+p-n} .$$

The above product is not greater than 0, because among the factors $$k+p-1,k+p-2,\ldots,k+p-(n-1)$$ there are exactly $$n-1-k$$ negative ones.

Similarly, the function $$f(x)=x^{p}$$ for $$p\in(-1,0)$$ satisfies property (2.5) for the set $$S=\{ 1,2,\ldots,{n-1}\}$$, because $$p<0$$ and among the factors $$k+p-1,k+p-2,\ldots,k+p-(n-1)$$ there are exactly $$n-k$$ negative ones. On the other hand, property (2.5) is not true for $$k=0$$.

Thus, like above, we have the following.

### Corollary 2.11

Assume that $$a,b\in(0,\infty)^{n}$$ be such that $$a\preceq b$$ and $$a_{1}a_{2}\cdots a_{n}=b_{1}b_{2}\cdots b_{n}$$. If $$p\in(-1,1)$$, then

$$a_{1}^{p}+a_{2}^{p}+ \cdots+a_{n}^{p}\leq b_{1}^{p}+b_{2}^{p}+ \cdots+b_{n}^{p} .$$

This inequality fails for $$-1< p<0$$ (but remains true for $$0< p<1$$) if the constraint $$a_{1}a_{2}\cdots a_{n}=b_{1}b_{2}\cdots b_{n}$$ is replaced by the weaker one $$a_{1}a_{2}\cdots a_{n}\leq b_{1}b_{2}\cdots b_{n}$$.

### Proof of Theorem 2.3

If S is empty, then $$E_{k}(a)=E_{k}(b)$$ for all $$k\in\{0,1,\ldots,n-1\}$$ and hence $$a=b$$, which immediately implies the inequality. We therefore assume that S is nonempty.

Let $$y:[0,1]\to\Delta_{n}$$ be the curve connecting points a and b as in Definition 2.2. Consider the function

\begin{aligned} \begin{aligned}[b] p(t,x)&=\bigl(x+y_{1}(t)\bigr) \bigl(x+y_{2}(t) \bigr)\cdots\bigl(x+y_{n}(t)\bigr)=\sum_{k=0}^{n-1} x^{k}E_{k}\bigl(y(t)\bigr) + x^{n} \\ &=(x+a_{1}) (x+a_{2})\cdots(x+a_{n})+\sum _{k\in S} x^{k} A_{k}(t) , \end{aligned} \end{aligned}
(2.11)

where $$A_{k}(t)=E_{k}(y(t))-E_{k}(a)$$ is a nondecreasing mapping. Our goal is to show that the function

$$\eta(t)=\sum_{i=1}^{n} f \bigl(y_{i}(t)\bigr)$$
(2.12)

is nondecreasing on $$[0,1]$$, i.e. we show that $$\eta'(t)\geq 0$$ a.e. on $$(0,1)$$.

To this end, fix $$i\in\{1,2,\ldots,n\}$$. Since $$p(t,-y_{i}(t))=0$$ for all $$t\in(0,1)$$, we obtain

$$\partial_{1} p\bigl(t,-y_{i}(t)\bigr)+\partial_{2} p\bigl(t,-y_{i}(t)\bigr)\cdot\bigl(-y_{i}'(t) \bigr)=0$$

for all $$t\in(0,1)$$ and therefore

$$\sum_{k\in S} \bigl(-y_{i}(t) \bigr)^{k} A'_{k}(t)+\prod _{j\neq i} \bigl(y_{j}(t)-y_{i}(t)\bigr) \cdot \bigl(-y_{i}'(t)\bigr)=0 ,$$
(2.13)

which gives

$$y_{i}'(t)=\sum_{k\in S} \bigl(-y_{i}(t)\bigr)^{k} A'_{k}(t) \biggl(\prod_{j\neq i} \bigl(y_{j}(t)-y_{i}(t) \bigr) \biggr)^{-1} .$$

This equality holds, if $$y_{i}(t)\neq y_{j}(t)$$ for $$i\neq j$$, which is true for all but countably many values of $$t\in(0,1)$$. For those values of t we get

\begin{aligned} \eta'(t)&=\sum_{i=1}^{n} f'\bigl(y_{i}(t)\bigr)\cdot y_{i}'(t) \\ &=\sum_{i=1}^{n} f' \bigl(y_{i}(t)\bigr)\cdot\sum_{k\in S} \bigl(-y_{i}(t)\bigr)^{k} A'_{k}(t) \biggl(\prod_{j\neq i} \bigl(y_{j}(t)-y_{i}(t) \bigr) \biggr)^{-1} \\ &=\sum_{k\in S} A'_{k}(t)\sum _{i=1}^{n} f' \bigl(y_{i}(t)\bigr)\cdot\bigl(-y_{i}(t)\bigr)^{k} \biggl(\prod_{j\neq i} \bigl(y_{j}(t)-y_{i}(t) \bigr) \biggr)^{-1} . \end{aligned}
(2.14)

Fix $$t\in(0,1)$$ such that $$y_{i}(t)\neq y_{j}(t)$$ for $$i\neq j$$ and write $$y_{i}=y_{i}(t)$$ for simplicity. Since $$A'_{k}(t)\geq0$$, we will be done if we show that

$$\widehat{D}:=\sum_{i=1}^{n} f'(y_{i})\cdot(-y_{i})^{k} \biggl( \prod_{j\neq i} (y_{j}-y_{i}) \biggr)^{-1}\geq0 \quad \mbox{for all }k\in S.$$

To this end, consider the polynomial

$$g(x)=\sum_{i=1}^{n} f'(y_{i}) \cdot(-y_{i})^{k} \biggl(\prod_{j\neq i} (y_{j}-y_{i}) \biggr)^{-1}\cdot\prod _{j\neq i} (x-y_{j}) .$$

The degree of g equals $$n-1$$ and the coefficient at $$x^{n-1}$$ is equal to $$\widehat{D}$$. Moreover,

$$g(y_{i})= f'(y_{i})\cdot(-y_{i})^{k} \cdot(-1)^{n-1} \quad (i=1,2,\ldots,n) .$$

Therefore the function $$h(x)=g(x)+(-1)^{n+k}x^{k}f'(x)$$ has n different roots $$y_{1},y_{2},\ldots,y_{n}$$ in the interval I. It follows that the function

$$h^{(n-1)}(x) =(n-1)! \widehat{D} + (-1)^{n+k} \bigl(x^{k}f'(x)\bigr)^{(n-1)}$$
(2.15)

has a root in the interval I, and since $$(-1)^{n+k}(x^{k}f'(x))^{(n-1)}\leq0$$ for all $$x\in I$$, it follows that $$\widehat{D}\geq0$$, which completes the proof of Theorem 2.3. □

### Proof of Theorem 2.4

Suppose, to the contrary, that $$(-1)^{k+n}(x^{k}f'(x))^{(n-1)}>0$$ for some $$x\in I$$ and some $$k\in S$$. Then $$(-1)^{k+n}(x^{k}f'(x))^{(n-1)}>0$$ holds for all x belonging to some interval J contained in I. Choose the numbers $$a_{1}< a_{2}<\cdots<a_{n}$$ from J and consider

$$p(t,x)=(x+a_{1})\cdot(x+a_{2})\cdot\,\cdots\, \cdot(x+a_{n})+t x^{k} .$$

Then for all sufficiently small t ($$0< t<\varepsilon$$), there exist different numbers $$y_{i}(t)$$ belonging to J, such that

$$p(t,x)=\bigl(x+y_{1}(t)\bigr) \bigl(x+y_{2}(t)\bigr)\cdots \bigl(x+y_{n}(t)\bigr) .$$

Then

$$x^{n} + \sum_{i=0}^{n-1} E_{i}(a) \cdot x^{i} + t x^{k} = p(t,x) = x^{n} + \sum_{i=0}^{n-1} E_{i}\bigl(y(t)\bigr) \cdot x^{i} ,$$

and since $$t>0$$, we see that a and $$b=y(t)$$ satisfy (2.8). We will be done if we show that

$$f(a_{1})+f(a_{2})+\cdots+f(a_{n})>f \bigl(y_{1}(t)\bigr)+f\bigl(y_{2}(t)\bigr)+\cdots+f \bigl(y_{n}(t)\bigr) .$$

We proceed in the same way as in the proof of Theorem 2.3. We define

$$\eta(t)=\sum_{i=1}^{n} f \bigl(y_{i}(t)\bigr)\quad \mbox{for }0< t<\varepsilon$$

and this time we want to show that $$\eta'(t)<0$$ for $$0< t<\varepsilon$$.

By the inverse mapping theorem (see the proof of Proposition 3.4 below for a more detailed explanation), $$y\in C^{1}(0,\varepsilon)$$ and therefore

$$\eta'(t)=\sum_{i=1}^{n} f'\bigl(y_{i}(t)\bigr)\cdot y_{i}'(t)= \sum_{i=1}^{n} f' \bigl(y_{i}(t)\bigr)\cdot\bigl(-y_{i}(t)\bigr)^{k} \biggl(\prod_{j\neq i} \bigl(y_{j}(t)-y_{i}(t) \bigr) \biggr)^{-1} .$$
(2.16)

Now, like previously, write $$y_{i}=y_{i}(t)$$ for simplicity. Our goal is therefore to prove that

$$\widehat{D}:=\sum_{i=1}^{n} f'(y_{i})\cdot(-y_{i})^{k} \biggl( \prod_{j\neq i} (y_{j}-y_{i}) \biggr)^{-1}< 0 .$$

Consider the polynomial

$$g(x)=\sum_{i=1}^{n} f'(y_{i}) \cdot(-y_{i})^{k} \biggl(\prod_{j\neq i} (y_{j}-y_{i}) \biggr)^{-1}\cdot\prod _{j\neq i} (x-y_{j}) .$$

The degree of g equals $$n-1$$ and the coefficient at $$x^{n-1}$$ is equal to $$\widehat{D}$$. Moreover, the function $$h(x)=g(x)+(-1)^{n+k}x^{k}f'(x)$$ has n different roots $$y_{1},y_{2},\ldots,y_{n}$$ in the interval J. It follows that the function

$$h^{(n-1)}(x) =(n-1)! \widehat{D} + (-1)^{n+k} \bigl(x^{k}f'(x)\bigr)^{(n-1)}$$

has a root in the interval J. Since $$(-1)^{n+k}(x^{k}f'(x))^{(n-1)}>0$$ for all $$x\in J$$, it follows that $$\widehat{D}<0$$, which completes the proof of Theorem 2.4. □

## 3 Construction of the connecting curve

In this section we prove that condition (2.4) implies $$a\preceq b$$, if $$2\leq n\leq4$$, $$I=(0,\infty)$$ and $$S\subseteq\{1,2,\ldots,n-1\}$$. However, we start with a construction of the desired curve for a general interval I, integer $$n\geq2$$ and set $$S\subseteq\{0,1,\ldots,n-1\}$$.

For $$a,b\in\Delta_{n}$$, we say that $$a< b$$, if $$a\neq b$$ and $$E_{k}(a)\leq E_{k}(b)$$ for all $$k=0,1,\ldots,n-1$$. We say that $$a\leq b$$, if $$a< b$$ or $$a=b$$.

### Definition 3.1

For $$a< b$$ denote by $$\mathcal{C}(a,b)$$ the set of all piecewise differentiable (i.e. continuous and differentiable in all but at most countably many points) curves y in $$\Delta_{n}$$ satisfying:

1. (a)

the curve $$y(t)$$ starts at a (i.e. $$y(0)=a$$, if the curve $$y(t)$$ is parametrised by the interval $$[0,\varepsilon]$$);

2. (b)

$$y(t)\in\operatorname{int}(\Delta_{n})$$ for all but at most countable many values t;

3. (c)

the mappings $$E_{k}(y(t))$$ are nondecreasing in t and $$E_{k}(y(t))\leq E_{k}(b)$$ for all t and each $$k=0,1,\ldots,n-1$$.

Note that a curve in $$\mathcal{C}(a,b)$$ does not necessarily end at the point b.

### Proposition 3.2

Let $$n\geq2$$ be a positive integer and let S be a nonempty subset of $$\{0,1,\ldots,n-1\}$$. Let, moreover, $$a,b\in\Delta_{n}$$ be such that (2.4) holds. Furthermore, suppose that for all $$c\in\Delta_{n}$$ with $$a\leq c< b$$ the set $$\mathcal{C}(c,b)$$ is nonempty. Then $$a\preceq b$$.

### Proof

Each element (curve) of $$\mathcal{C}(a,b)$$ is a (closed) subset of $$\Delta_{n}$$. We equip the set $$\mathcal{C}(a,b)$$ with the inclusion relation , obtaining a nonempty partially ordered set $$(\mathcal{C}(a,b),\subseteq)$$. We are going to show that each chain $$\{y_{i}\}_{i\in\mathcal{I}}$$ has an upper bound in $$\mathcal{C}(a,b)$$.

To achieve this, consider the curve

$$y_{0}=\overline{\bigcup_{i\in\mathcal{I}} y_{i}} ,$$

i.e. the concatenation of the curves $$y_{i}$$. Then obviously $$y_{0}$$ satisfies conditions (a) and (c) of Definition 3.1. To prove (b) assume that $$y_{0}$$ is parametrised on $$[0,1]$$. Then for each positive integer k the curve $$y_{k}$$, defined as the restriction of $$y_{0}$$ to the interval $$[0,1-{1\over k}]$$, is contained in some curve $$y_{i}\in\mathcal{C}(a,b)$$ of the given chain $$\{y_{i}\}$$. Therefore $$y_{k}(t)$$ is piecewise differentiable and satisfies condition (b) for each positive integer k. Moreover,

$$y_{0}=\overline{\bigcup_{k=1}^{\infty}y_{k}} .$$

Hence $$y_{0}$$ is piecewise differentiable and satisfies (b) as well.

Now, by the Kuratowski-Zorn lemma, there exists a maximal element y in $$(\mathcal{C}(a,b),\subseteq)$$. We show that y is a desired curve connecting the points a and b, which will imply that $$a\preceq b$$.

To this end, it is enough to show that, if the curve y is parametrised on $$[0,1]$$, then $$y(1)=b$$. Suppose, to the contrary, that $$y(1)=c\neq b$$. Then $$a\leq c< b$$, and hence the set $$\mathcal{C}(c,b)$$ is nonempty. Thus the curve y can be extended beyond the point c, which contradicts the fact that y is a maximal element in $$\mathcal{C}(a,b)$$. This completes the proof of Proposition 3.2. □

From now on assume that $$I=(0,\infty)$$ and S is a nonempty subset of $$\{1,2,\ldots,n-1\}$$.

In order to prove that (2.4) implies $$a\preceq b$$, it suffices to show that the sets $$\mathcal{C}(a,b)$$ for $$a,b\in\Delta_{n}$$ with $$a< b$$ are nonempty. This is implied by the following conjecture, which we will prove later for $$n\leq4$$.

### Conjecture 3.3

Let $$n\geq2$$ be an integer and $$a\in\Delta_{n}$$. Let S be a nonempty subset of $$\{1,2,\ldots,n-1\}$$ with the property that there exist $$A_{k}>0$$ for $$k\in S$$ such that all the roots of the polynomial

$$q(x)=(x+a_{1}) (x+a_{2})\cdots(x+a_{n})+\sum _{k\in S} A_{k}x^{k}$$

are real (and hence negative). Then there exist mappings $$B_{k}:[0,\varepsilon ]\to\mathbb{R}$$ ($$k\in S$$) continuous on $$[0,\varepsilon]$$, differentiable on $$(0,\varepsilon)$$, and nondecreasing with $$B_{k}(0)=0$$ such that $$\sum_{k\in S} B_{k}(t)$$ is increasing on $$[0,\varepsilon]$$ and for all sufficiently small values of $$t>0$$ the polynomial

$$(x+a_{1}) (x+a_{2})\cdots(x+a_{n})+\sum _{k\in S} B_{k}(t)x^{k}$$

has n distinct real (and hence negative) roots.

Now we show how Conjecture 3.3 implies that the sets $$\mathcal{C}(a,b)$$ are nonempty.

### Proposition 3.4

Let n and S be such that the conjecture holds. Let, moreover, $$a,b\in\Delta_{n}$$ be such that (2.4) holds. Then the set $$\mathcal{C}(a,b)$$ is nonempty.

### Proof

Consider the polynomials

$$p(x)=(x+a_{1}) (x+a_{2})\cdots(x+a_{n}) \quad \mbox{and}\quad q(x)=(x+b_{1}) (x+b_{2})\cdots(x+b_{n}) .$$

Then

$$q(x)-p(x)=\sum_{k=0}^{n-1} \bigl(E_{k}(b)-E_{k}(a)\bigr)x^{k}=\sum _{k\in S} A_{k}x^{k} ,$$

where $$A_{k}>0$$ for all $$k\in S$$. According to the conjecture, there exist nondecreasing mappings $$B_{k}:[0,\varepsilon]\to\mathbb{R}$$, continuous on $$[0,\varepsilon]$$ and differentiable on $$(0,\varepsilon)$$, with $$B_{k}(0)=0$$, such that $$\sum_{k\in S}B_{k}(t)$$ is increasing on $$[0,\varepsilon]$$ and for all $$t\in(0,\varepsilon)$$ the polynomial

$$p(x)+ \sum_{k\in S} B_{k}(t)x^{k}$$

has n distinct real (and hence negative) roots $$-y_{n}(t)<-y_{n-1}(t)<\cdots<-y_{1}(t)<0$$. We show that $$y(t)=(y_{1}(t),y_{2}(t),\ldots,y_{n}(t))$$ defines a differentiable curve (parametrised on $$[0,\varepsilon]$$) that belongs to $$\mathcal{C}(a,b)$$, provided ε is chosen in such a way that $$B_{k}(\varepsilon)\leq A_{k}$$ for $$k\in S$$.

Consider the mapping $$\Psi :\overline{\Delta_{n}}\to\Psi (\overline{\Delta_{n}})$$ given by

$$\Psi(y)=\bigl(E_{n-1}(y),E_{n-2}(y),\ldots,E_{0}(y) \bigr) .$$

Then it follows from Remark 1.3 that the mapping Ψ is injective, hence Ψ is a continuous bijection defined on a closed subset of $$\mathbb{R}^{n}$$. Therefore the restriction $$\Psi|_{U}$$ of Ψ to a neighbourhood U of a is continuously invertible and thus

$$y(t)=\Psi^{-1}\bigl(\Psi(a)+\bigl(B_{0}(t),B_{1}(t), \ldots,B_{n-1}(t)\bigr)\bigr) \quad \bigl(t\in[0,\varepsilon]\bigr)$$

(here we put $$B_{k}(t)=0$$ for $$k\notin S$$) is a curve starting at a; note that $$\Psi(a)+(B_{0}(t),B_{1}(t), \ldots ,B_{n-1}(t))$$ is contained in $$\Psi(U)$$ for sufficiently small ε. Moreover $$y(t)\in\Delta_{n}$$. Hence condition (a) is satisfied. Since $$y(t)\in\operatorname{int}(\Delta_{n})$$ for all $$t\in(0,\varepsilon)$$, condition (b) holds. It is also clear that (c) is satisfied, since $$E_{k}(y(t))=E_{k}(a)+B_{k}(t)\leq E_{k}(a)+A_{k}=E_{k}(b)$$ for all $$k\in\{ 0,1,\ldots,n-1\}$$.

It remains to prove that $$y(t)$$ is differentiable on $$(0,\varepsilon)$$. This, however, is a consequence of the inverse mapping theorem, if we show that

$$\operatorname{det}\bigl[D\Psi(y)\bigr]\neq0\quad \mbox{for all }y\in \operatorname{int}(\Delta_{n}).$$

To this end, let $$V(y)$$ be the $$n\times n$$ Vandermonde-type matrix given by $$V_{ij}(y)=(-y_{i})^{n-j}$$ ($$1\leq i,j\leq n$$). This matrix is obtained from the standard Vandermonde matrix

$$W (-y_{1}, -y_{2}, \ldots, -y_{n}) = \begin{pmatrix} 1 & -y_{1} & (-y_{1})^{2} & \cdots& (-y_{1})^{n-1} \\ 1 & -y_{2} & (-y_{2})^{2} & \cdots& (-y_{2})^{n-1} \\ 1 & -y_{3} & (-y_{3})^{2} & \cdots& (-y_{3})^{n-1} \\ \vdots& \vdots& \vdots& \ddots& \vdots \\ 1 & -y_{n} & (-y_{n})^{2} & \cdots& (-y_{n})^{n-1} \end{pmatrix}$$
(3.1)

by reversing the order of columns of W.

Since 

$$\bigl(D\Psi(y)\bigr)_{jk} = \frac{\partial}{\partial y_{k}} E_{n-j}(y) = \begin{cases} 1, & j=1 , \\ E_{n-j}(y^{(k)}), & j>1 , \end{cases}$$

where $$y^{(k)} = (y_{1},\ldots,y_{k-1},y_{k+1},\ldots,y_{n})$$ is y with its kth component removed, it follows from the general formula

$$t^{n-1}+\sum_{j=0}^{n-2}t^{j}E_{j}(z_{1},z_{2}, \ldots ,z_{n-1})=(t+z_{1}) (t+z_{2}) \cdots(t+z_{n-1})$$
(3.2)

that

\begin{aligned} \bigl(V(y) \cdot D\Psi(y)\bigr)_{ik} &= \sum _{j=1}^{n} \bigl(V(y)\bigr)_{ij} \cdot \bigl(D\Psi (y)\bigr)_{jk} \\ &= (-y_{i})^{n-1} + \sum_{j=2}^{n} (-y_{i})^{n-j} \cdot E_{n-j}\bigl(y^{(k)} \bigr) \\ &= (-y_{i})^{n-1} + \sum_{j=0}^{n-2} (-y_{i})^{j} \cdot E_{j}\bigl(y^{(k)} \bigr)= \prod_{j\neq k} (y_{j}-y_{i}) \end{aligned}

and thus

$$V(y)\cdot D\Psi(y)= \operatorname{diag} \biggl(\prod _{j\neq1}(y_{j}-y_{1}),\prod _{j\neq2}(y_{j}-y_{2}),\ldots ,\prod _{j\neq n}(y_{j}-y_{n}) \biggr) .$$
(3.3)

It is well known that

$$\operatorname{det}\bigl[V(y)\bigr]=\prod_{i< j}(y_{j}-y_{i}) \neq0 \quad (y\in\operatorname{int} {\Delta_{n}}) .$$

Therefore we obtain

$$\operatorname{det}\bigl[D\Psi(y)\bigr]=\prod_{i< j}(y_{i}-y_{j}) \neq0 \quad (y\in\operatorname{int} {\Delta_{n}}) ,$$

which completes the proof of Proposition 3.4. □

### Lemma 3.5

Assume that $$n\geq3$$ is odd and let $$0< a_{1}\leq a_{2}\leq\cdots\leq a_{n}$$. Let, moreover, $$A_{k}\geq0$$ for $$k=1,2,\ldots,(n-1)/2$$ with at least one $$A_{k}$$ not equal to 0. Consider the polynomials

\begin{aligned} &P(x)=(x+a_{1}) (x+a_{2}) \cdots(x+a_{n})+\sum_{k=1}^{(n-1)/2} A_{k}x^{2k-1}, \\ &Q(x)=(x+a_{1}) (x+a_{2})\cdots(x+a_{n})+\sum _{k=1}^{(n-1)/2} A_{k}x^{2k} . \end{aligned}
(3.4)

Then the polynomial P has exactly one root in the interval $$(-a_{1},0)$$ and at most two roots in the interval $$(-a_{n},-a_{n-1})$$. Moreover, the polynomial Q has exactly one root in the interval $$(-\infty,-a_{n})$$ and at most two roots in the interval $$(-a_{2},-a_{1})$$.

### Proof

That P has exactly one root in $$(-a_{1},0)$$ follows immediately from the observation that $$P(-a_{1})<0$$, $$P(0)>0$$ and $$P'(x)>0$$ on $$(-a_{1},0)$$.

Now we show that Q has exactly one root in $$(-\infty,-a_{n})$$.

Dividing the equation $$Q(x)=0$$ by $$x^{n}a_{1}a_{2}\cdots a_{n}$$ and substituting $$z=1/x$$ and $$b_{i}=1/a_{i}$$ yield the equation $$P_{0}(z)=0$$, where

$$P_{0}(z)=(z+b_{1}) (z+b_{2})\cdots(z+b_{n})+ \sum_{k=1}^{(n-1)/2} B_{k}z^{2k-1}$$

for some nonnegative numbers $$B_{k}$$, not all equal to 0. We already know that $$P_{0}$$ has exactly one root in the interval $$(-b_{n},0)$$, so it follows that Q has exactly one root in the interval $$(-\infty,-a_{n})$$.

Now we prove that Q has at most two roots in the interval $$(-a_{2},-a_{1})$$. To the contrary, suppose that Q has at least three roots in $$(-a_{2},-a_{1})$$. Since $$Q(-a_{2})>0$$ and $$Q(-a_{1})>0$$, it follows that Q has an even number, and hence at least four, roots in the interval $$(-a_{2},-a_{1})$$.

Let $$0>-c_{1}\geq-c_{2}\geq\cdots\geq-c_{n-1}$$ be the roots of $$p'(x)=0$$, where

$$p(x)=(x+a_{1}) (x+a_{2})\cdots(x+a_{n}) .$$
(3.5)

Then $$a_{1}< c_{1}<a_{2}$$. The polynomial $$Q(x)$$ is decreasing on the interval $$[-a_{2},-c_{1}]$$, so it has at most one root in this interval. Therefore the polynomial Q has at least three roots in the interval $$(-c_{1},-a_{1})$$, and consequently the equation $$Q''(x)=0$$ has a root in $$(-c_{1},-a_{1})$$. But $$Q''(x)>0$$ for all $$x>-c_{1}$$, a contradiction. Hence Q must have at most two roots in $$(-a_{2},-a_{1})$$.

Finally, to prove that P has at most two roots in the interval $$(-a_{n},-a_{n-1})$$, divide the equation $$P(x)=0$$ by $$x^{n}a_{1}a_{2}\cdots a_{n}$$ and substitute $$z=1/x$$ and $$b_{i}=1/a_{i}$$. This reduces to the equation $$Q_{0}(z)=0$$, where

$$Q_{0}(z)=(z+b_{1}) (z+b_{2})\cdots(z+b_{n})+ \sum_{k=1}^{(n-1)/2} B_{k}z^{2k}$$

for some nonnegative numbers $$B_{k}$$, not all equal to 0. We already know that $$Q_{0}$$ has at most two roots in the interval $$(-b_{n-1},-b_{n})$$, so it follows that P has at most two roots in the interval $$(-a_{n},-a_{n-1})$$. This completes the proof of Lemma 3.5. □

The same proof yields an analogous result for even values of n.

### Lemma 3.6

Assume that $$n\geq2$$ is even and let $$0< a_{1}\leq a_{2}\leq\cdots\leq a_{n}$$. Let, moreover, $$A_{k}\geq0$$ for $$k=1,2,\ldots,n/2$$ and not all of the $$A_{k}$$ ’s are equal to 0. Consider the polynomials

\begin{aligned} &P(x)=(x+a_{1}) (x+a_{2}) \cdots(x+a_{n})+\sum_{k=1}^{n/2} A_{k}x^{2k-1}, \\ &Q(x)=(x+a_{1}) (x+a_{2})\cdots(x+a_{n})+\sum _{k=1}^{n/2-1} A_{k}x^{2k} . \end{aligned}
(3.6)

Then the polynomial P has exactly one root in each of the intervals $$(-\infty,-a_{n})$$ and $$(-a_{1},0)$$ and Q has at most two roots in each of the intervals $$(-a_{n},-a_{n-1})$$ and $$(-a_{2},-a_{1})$$.

### Proof

The same proof as that for Lemma 3.5 can be used. □

Now we turn to the proof of Conjecture 3.3 for $$2\leq n\leq 4$$ and an arbitrary nonempty set $$S\subseteq\{1,2,\ldots,n-1\}$$.

We first make some useful general remarks.

Let $$I(a)=\{i\in\{1,2,\ldots,n-1\} : a_{i}=a_{i+1}\}$$. If $$I(a)$$ is empty, then the conjecture holds. Indeed, if $$k\in S$$, then all the roots of the polynomial

$$(x+a_{1}) (x+a_{2})\cdots(x+a_{k})+t x^{k}$$

are, for all sufficiently small $$t>0$$, real and distinct.

On the other hand, if $$I(a)=\{1,2,\ldots,n-1\}$$, then only the set $$S=\{1,2,\ldots,n-1\}$$ possibly satisfies the assumptions of the conjecture. Indeed, suppose that $$l\notin S$$ and let $$-b_{1}\geq-b_{2}\geq\cdots\geq-b_{n}$$ be the roots of

$$q(x)=(x+a_{1})^{n}+\sum_{k\in S} A_{k}x^{k} .$$

Then by the inequality of arithmetic and geometric means, we obtain

$${E_{l}(a)\over {n\choose l}}={E_{l}(b)\over {n\choose l}} \geq \bigl(E_{0}(b)\bigr)^{(n-l)/n}=\bigl(E_{0}(a) \bigr)^{(n-l)/n}={E_{l}(a)\over {n\choose l}} ,$$
(3.7)

and hence $$b_{1}=b_{2}=\cdots=b_{n}$$. Since $$E_{0}(a)=E_{0}(b)$$, it follows that $$a=b$$, i.e. $$A_{k}=0$$ for all $$k\in S$$. A contradiction.

Let I be a nonempty subset of $$\{1,2,\ldots,n-1\}$$. We observe that the conjecture is true for a set S and all $$a\in\Delta_{n}$$ with $$I(a)=I$$, if it is true for a set $$T=\{n-k : k\in S\}$$ and all $$b\in\Delta_{n}$$ with $$I(b)=\{n-i : i\in I\}$$. Indeed, if all the roots of the polynomial

$$q(x)=(x+a_{1}) (x+a_{2})\cdots(x+a_{n})+\sum _{k\in S} A_{k}x^{k}$$

are real, then substituting $$x=1/z$$ and $$a_{i}=1/b_{i}$$, we infer that all the roots of the polynomial

$$r(z)=(z+b_{1}) (z+b_{2})\cdots(z+b_{n})+\sum _{l\in T} B_{l}z^{l}$$

are real. Hence there exist mappings $$C_{l}(t)$$ with $$C_{l}(0)=0$$, continuous on $$[0,\varepsilon]$$, differentiable on $$(0,\varepsilon)$$ and nondecreasing such that the polynomial

$$(z+b_{1}) (z+b_{2})\cdots(z+b_{n})+\sum _{l\in T} C_{l}(t)z^{l}$$

has n distinct real roots. Substituting $$z=1/x$$ and $$b_{i}=1/a_{i}$$, we infer that the polynomial

$$(x+a_{1}) (x+a_{2})\cdots(x+a_{n})+\sum _{k\in S} C_{n-k}(t)x^{k}$$

has n distinct real roots.

For $$n=2$$ the only possibility for the set S is $$\{1\}$$ and it is enough to notice that the polynomial $$(x+a_{1})(x+a_{2})+t x$$ has two distinct real roots for any $$t>0$$.

Assume now $$n=3$$. Then, in view of the above remarks, we have to consider two cases: (1) $$a_{1}< a_{2}=a_{3}$$; (2) $$a_{1}=a_{2}=a_{3}$$.

(1) If $$2\notin S$$, then the condition of Conjecture 3.3 cannot be satisfied since for $$A_{1}>0$$, according Lemma 3.5, the polynomial

$$P(x) = (x+a_{1}) (x+a_{2})^{2}+A_{1}x$$

has only one real root in the interval $$(-a_{1},0)$$ and obviously no roots on $$\mathbb{R}\setminus(-a_{1},0)$$. Thus P has only one real root for all $$A_{1}>0$$. We can therefore assume $$2\in S$$, and for all sufficiently small $$t>0$$, the polynomial

$$(x+a_{1}) (x+a_{2})^{2}+t x^{2}$$

has three distinct real roots.

(2) According to the above remarks, $$S=\{1,2\}$$. Then the polynomial $$(x+a_{1})^{3}+t a_{1}x+t x^{2}$$ has three distinct real roots for all sufficiently small $$t>0$$.

Assume $$n=4$$. In this case we have five possibilities: (1) $$a_{1}=a_{2}< a_{3}<a_{4}$$; (2) $$a_{1}< a_{2}=a_{3}<a_{4}$$; (3) $$a_{1}< a_{2}=a_{3}=a_{4}$$; (4) $$a_{1}=a_{2}< a_{3}=a_{4}$$; (5) $$a_{1}=a_{2}=a_{3}=a_{4}$$.

(1) We note that $$S\neq\{2\}$$, since, by Lemma 3.6, the polynomial

$$Q(x)=(x+a_{1})^{2}(x+a_{3}) (x+a_{4})+A_{2}x^{2} \quad \mbox{for }A_{2}>0$$

has at most two real roots in the interval $$(-a_{4},-a_{3})$$ and obviously no roots on $$\mathbb{R}\setminus(-a_{4},-a_{3})$$. Thus Q has at most two real roots. Therefore S contains an odd integer k. Then for all sufficiently small $$t>0$$, the polynomial $$(x+a_{1})^{2}(x+a_{3})(x+a_{4})+t x^{k}$$ has four distinct real roots.

(2) Note that $$2\in S$$, since by Lemma 3.6, the polynomial

$$(x+a_{1}) (x+a_{2})^{2}(x+a_{4})+A_{1}x+A_{3}x^{3} \quad \mbox{for }A_{1},A_{3}>0$$

has at most two real roots. Then for all sufficiently small $$t>0$$, the polynomial

$$(x+a_{1}) (x+a_{2})^{2}(x+a_{4})+t x^{2}$$

has four distinct real roots.

(3) We observe that $$\{1,2\}\subset S$$ or $$\{2,3\}\subset S$$, since by Lemma 3.6, each of the polynomials

$$(x+a_{1}) (x+a_{2})^{3}+A_{1}x+A_{3}x^{3} \quad \mbox{and}\quad (x+a_{1}) (x+a_{2})^{3}+A_{2}x^{2} \quad \mbox{for }A_{1},A_{2},A_{3}>0$$

as well as

$$(x+a_{1}) (x+a_{2})^{3}+A_{1}x\quad \mbox{and}\quad (x+a_{1}) (x+a_{2})^{3}+A_{3}x^{3} \quad \mbox{for }A_{1},A_{3}>0$$

has at most two real roots. Moreover, we prove that $$S\neq\{1,2\}$$.

Suppose that the polynomial $$Q(x)=(x+a_{1})(x+a_{2})^{3}+A_{1}x+A_{2}x^{2}$$ has four real roots. Let $$Q_{1}(x)=(x+a_{1})(x+a_{2})^{3}$$ and $$Q_{2}(x)=A_{1}x+A_{2}x^{2}$$. Let $$-c\neq a_{2}$$ be the root of the polynomial $$Q_{1}'(x)$$ and let −d be the root of $$Q_{2}'(x)$$.

If $$d< c$$, then Q is decreasing on $$(-\infty,-c]$$, so Q has at most one root in this interval. Therefore Q has at least three roots in the interval $$(-c,0)$$. Thus $$Q''(x)$$ has a root in the interval $$(-c,0)$$, which is impossible, since $$Q''(x)>0$$ on $$(-c,0)$$.

If $$a_{2}\geq d\geq c$$, then Q is increasing on the interval $$[-c,0)$$ and decreasing on the interval $$(-\infty,-d]$$, so Q must have at least two roots in the interval $$(-d,-c)$$. But $$Q(x)<0$$ on this interval.

Finally, if $$d>a_{2}$$, then Q may only have roots in the union $$(-\infty,a_{2})\cup(-a_{1},0)$$. But Q is increasing on $$(-a_{1},0)$$, so Q has three roots in $$(-\infty,a_{2})$$. This, however, is impossible, since $$Q''(x)>0$$ for $$x\in(-\infty,a_{2})$$. Thus $$\{2,3\}\subseteq S$$ and the polynomial

$$(x+a_{1}) (x+a_{2})^{3}+t x^{2}(x+a_{2})$$

has, for all sufficiently small $$t>0$$, four distinct roots.

(4) Since the polynomial $$(x+a_{1})^{2}(x+a_{3})^{2}+A_{2}x^{2}$$ has no real roots, $$1\in S$$ or $$3\in S$$. Then the polynomial $$(x+a_{1})^{2}(x+a_{3})^{2}+t x^{k}$$ for $$k=1,3$$ has, for all sufficiently small $$t>0$$, four distinct real roots.

(5) In view of the above remarks, $$S=\{1,2,3\}$$. Consider

$$r(x)=(x+a_{1})^{4}+t x^{3}+2 t a_{1}x^{2}+t\bigl(a_{1}^{2}-t^{2} \bigr)x=(x+a_{1})^{4}+t x\bigl((x+a_{1})^{2}-t^{2} \bigr) .$$

Then for all sufficiently small $$t>0$$, $$a_{1}^{2}-t^{2}>0$$, and the polynomial r has four distinct real roots, because

\begin{aligned}& r(-a_{1}-2t)=t^{3}(10t-3a_{1})< 0 , \qquad r(-a_{1})=a_{1}t^{3}>0 \quad \mbox{and} \\& r(-a_{1}+2t)=t^{3}(22 t-3a_{1})<0 . \end{aligned}

Thus we have proved the following.

### Corollary 3.7

Conjecture  3.3 is true if $$2\leq n\leq4$$ and S is an arbitrary nonempty subset of $$\{1,2,\ldots,n-1\}$$.

This implies that the sum of squared logarithms inequality (Conjecture 1.2) holds also for $$n=4$$.

### Corollary 3.8

(Sum of squared logarithms inequality for $$n=4$$)

Let $$a_{1},a_{2},a_{3},a_{4},b_{1},b_{2}, b_{3},b_{4}>0$$ be given positive numbers such that

\begin{aligned}& a_{1}+a_{2}+a_{3}+a_{4} \le b_{1}+b_{2}+b_{3}+b_{4} , \\& a_{1} a_{2}+a_{1} a_{3}+ a_{2} a_{3}+a_{1} a_{4}+a_{2} a_{4}+a_{3} a_{4} \le b_{1} b_{2}+b_{1} b_{3}+ b_{2} b_{3}+b_{1} b_{4}+b_{2} b_{4}+b_{3} b_{4} , \\& a_{1} a_{2} a_{3}+a_{1} a_{2} a_{4}+a_{2} a_{3} a_{4}+a_{1} a_{3} a_{4} \le b_{1} b_{2} b_{3}+b_{1} b_{2} b_{4}+b_{2} b_{3} b_{4}+b_{1} b_{3} b_{4} , \\& a_{1} a_{2} a_{3} a_{4} = b_{1} b_{2} b_{3} b_{4} . \end{aligned}

Then

$$\log^{2} a_{1}+\log^{2} a_{2}+ \log^{2} a_{3}+\log^{2} a_{4} \le \log^{2} b_{1}+\log ^{2} b_{2}+ \log^{2} b_{3}+\log^{2} b_{4} .$$

### Proof

Use Corollary 3.7 and observe that S may be an arbitrary subset of $$\{1,2,3\}$$. □

### Corollary 3.9

Let $$n\geq2$$ be an integer and let T be an arbitrary subset of $$\{1,2,\ldots, n-1\}$$. Assume that the Conjecture 3.3 holds for n and for any nonempty subset S of T. Let, moreover, $$f\in C^{n}(0,\infty)$$. Then the inequality

$$f(a_{1})+f(a_{2})+\cdots+f(a_{n})\leq f(b_{1})+f(b_{2})+\cdots+f(b_{n})$$

holds for all $$a,b\in\Delta_{n}$$ satisfying

$$E_{k}(a)\leq E_{k}(b) \quad \textit{for } k\in T \quad \textit{and}\quad E_{k}(a)=E_{k}(b) \quad \textit{for } k=0 \textit{ or }k\notin T$$
(3.8)

if and only if

$$(-1)^{n+k}\bigl(x^{k} f'(x) \bigr)^{(n-1)}\leq0 \quad \textit{for all }x>0 \textit{ and all } k\in T .$$
(3.9)

### Proof

Assume first (3.9) holds and let $$a,b\in\Delta_{n}$$ satisfy (3.8). Consider any $$c\in\Delta_{n}$$ with $$a\leq c< b$$. Then the pair c, b satisfies condition (2.4) for some nonempty subset S of T. Therefore by Proposition 3.4, the set $$\mathcal{C}(c,b)$$ is nonempty and hence by Proposition 3.2, $$a\preceq b$$. Now Theorem 2.3 implies that inequality (2.6) holds.

Conversely, if (2.6) holds for all $$a,b\in\Delta_{n}$$ satisfying (3.8), then (2.6) also holds for all $$a,b\in\Delta_{n}$$ satisfying condition (2.4) with $$S=T$$. Thus Theorem 2.4 implies (3.9). This completes the proof. □

## 4 Outlook

Our result generalises and extends previous results on the sum of squared logarithms inequality. Indeed, compared to the proof in  our development here views the problem from a different angle in that it is not the logarithm function that defines the problem, but a certain monotonicity property in the geometry of polynomials, explicitly stated in Conjecture 3.3.

If one tries to adopt the above proof of Conjecture 3.3 for $$n\leq4$$ to the case $$n\geq5$$, one has to deal with approximately $$2^{n}$$ cases considered separately. Therefore it is clear that the extension to natural numbers n beyond $$n=6$$, say, is out of reach with such a method. Instead, a general argument should be found to prove or disprove Conjecture 3.3 for general n. Furthermore, it might be worthwhile to develop a better understanding of the differential inequality condition $$(-1)^{n+k}(x^{k} f'(x))^{(n-1)}\leq0$$.

## References

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## Acknowledgements

We thank Johannes Lankeit (Universität Paderborn) as well as Robert Martin (Universität Duisburg-Essen) for their help in revising this paper.

## Author information

Authors

### Corresponding author

Correspondence to Waldemar Pompe.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

Both authors contributed fully to all parts of this paper. Both authors read and approved the final manuscript.

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