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 [1] 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 ([2]; see also [3]) as well as in nonlinear elasticity theory [4–7]. 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 [1].

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:

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

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

• (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 [8]

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

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 [1] 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 $(-1)^{n+k}(x^{k} f'(x))^{(n-1)}\leq0$.