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# An application of matrix inequalities to certain functional inequalities involving fractional powers

Journal of Inequalities and Applications20122012:221

https://doi.org/10.1186/1029-242X-2012-221

• Accepted: 20 September 2012
• Published:

## Abstract

We will show certain functional inequalities involving fractional powers, making use of the Furuta inequality and Tanahashi’s argument.

MSC:26D07, 26A09, 39B62, 47A63.

## Keywords

• inequalities
• fractional powers
• matrix inequalities
• Furuta inequality

## 1 Introduction

Let x be an arbitrary positive real number. One can easily see the inequality
$\left({x}^{\frac{3}{2}}-1\right)\left({x}^{2}-1\right)\le \frac{6}{5}\left({x}^{\frac{5}{2}}-1\right)\left(x-1\right),$
for instance, is reduced to a simple polynomial inequality by putting $t={x}^{\frac{1}{2}}$. However, at least to the author, it seems not easy to give an elementary proof of the inequality
${x}^{\frac{2-\sqrt{2}+\sqrt{3}}{4}}\left({x}^{\sqrt{2}}-1\right)\left({x}^{\frac{\sqrt{2}+\sqrt{3}}{2}}-1\right)\le \frac{1}{\sqrt{2}}\left({x}^{\sqrt{2}+\sqrt{3}}-1\right)\left(x-1\right),$

which has a very similar form to the preceding one although their corresponding numerical parts are different.

Theorem 1.1 Let $0\le p$, $1\le q$ and $0\le r$ with $p+r\le \left(1+r\right)q$. If $0, then
${x}^{\frac{1+r-\frac{p+r}{q}}{2}}\left({x}^{p}-1\right)\left({x}^{\frac{p+r}{q}}-1\right)\le \frac{p}{q}\left({x}^{p+r}-1\right)\left(x-1\right).$
(1)

An elementary approach to proving the inequality (1) might be to consider the power series expansion.

Put $t=x-1$, $c=\frac{1+r-\frac{p+r}{q}}{2}$ and
$f\left(t\right)=\frac{p}{q}\left({\left(1+t\right)}^{p+r}-1\right)t-{\left(1+t\right)}^{c}\left({\left(1+t\right)}^{p}-1\right)\left({\left(1+t\right)}^{\frac{p+r}{q}}-1\right).$
Then we can expand $f\left(t\right)$ around $t=0$ as
$\begin{array}{rcl}f\left(t\right)& =& \frac{p}{q}\left\{p+r+\left(\genfrac{}{}{0}{}{p+r}{2}\right)t+\left(\genfrac{}{}{0}{}{p+r}{3}\right){t}^{2}+\left(\genfrac{}{}{0}{}{p+r}{4}\right){t}^{3}+\left(\genfrac{}{}{0}{}{p+r}{5}\right){t}^{4}+\cdots \right\}\cdot {t}^{2}\\ -\left\{1+ct+\left(\genfrac{}{}{0}{}{c}{2}\right){t}^{2}+\left(\genfrac{}{}{0}{}{c}{3}\right){t}^{3}+\left(\genfrac{}{}{0}{}{c}{4}\right){t}^{4}+\cdots \right\}\\ \cdot \left\{p+\left(\genfrac{}{}{0}{}{p}{2}\right)t+\left(\genfrac{}{}{0}{}{p}{3}\right){t}^{2}+\left(\genfrac{}{}{0}{}{p}{4}\right){t}^{3}+\left(\genfrac{}{}{0}{}{p}{5}\right){t}^{4}+\cdots \right\}\\ \cdot \left\{\frac{p+r}{q}+\left(\genfrac{}{}{0}{}{\frac{p+r}{q}}{2}\right)t+\left(\genfrac{}{}{0}{}{\frac{p+r}{q}}{3}\right){t}^{2}+\left(\genfrac{}{}{0}{}{\frac{p+r}{q}}{4}\right){t}^{3}+\left(\genfrac{}{}{0}{}{\frac{p+r}{q}}{5}\right){t}^{4}+\cdots \right\}\cdot {t}^{2}\\ =& {a}_{4}{t}^{4}+{a}_{5}{t}^{5}+{a}_{6}{t}^{6}+\cdots .\end{array}$
Thus, the constant term and the coefficients of t, ${t}^{2}$ and ${t}^{3}$ are 0. Further, one can obtain
$\begin{array}{c}{a}_{4}=\frac{p\left(p+r\right)}{24q}\left({r}^{2}+2pr+1-{\left(\frac{p+r}{q}\right)}^{2}\right),\hfill \\ {a}_{5}=\frac{p\left(p+r\right)\left(p+r-3\right)}{48q}\left({r}^{2}+2pr+1-{\left(\frac{p+r}{q}\right)}^{2}\right)\hfill \end{array}$
and
$\begin{array}{rcl}{a}_{6}& =& -\frac{p\left(p+r\right)}{5760q}\left\{3{\left(\frac{p+r}{q}\right)}^{4}+10{\left(\frac{p+r}{q}\right)}^{2}\left\{3\left(p+r\right)\left(p+r-8\right)+{p}^{2}+41\right\}\\ -33{\left(p+r\right)}^{4}+240{\left(p+r\right)}^{3}+30{\left(p+r\right)}^{2}\left({p}^{2}-15\right)-240\left(p+r\right)\left({p}^{2}-1\right)\\ +\left(3{p}^{2}+413\right)\left({p}^{2}-1\right)\right\}.\end{array}$

Thus, if the assumption for the parameters p, q and r in Theorem 1.1 is satisfied, then we have $0<{a}_{4}$. However, the signature of ${a}_{5}$ and ${a}_{6}$ depends on parameters, and one cannot see any signs of a simple rule among the coefficients of higher order terms. Although $f\left(t\right)$ is non-negative on a sufficiently small neighborhood of $t=0$, it seems difficult to show that $f\left(t\right)$ is non-negative entirely on $-1 by such an argument as above.

Let us recall some fundamental concepts on related matrix inequalities. A capital letter means a matrix whose entries are complex numbers. A square matrix T is said to be positive semidefinite (denoted by $0\le T$) if $0\le \left(Tx,x\right)$ for all vectors x. We write $0 if T is positive semidefinite and invertible. For two selfadjoint matrices ${T}_{1}$ and ${T}_{2}$ of the same size, a matrix inequality ${T}_{1}\le {T}_{2}$ is defined by $0\le {T}_{2}-{T}_{1}$.

The celebrated Löwner-Heinz theorem includes:

Theorem 1.2 [1, 2]

Let $0\le p\le 1$. If $0\le B\le A$, then ${B}^{p}\le {A}^{p}$.

For $1, $0\le B\le A$ does not always ensure ${B}^{p}\le {A}^{p}$. Furuta obtained an epoch-making extension of the Löwner-Heinz inequality by using the Löwner-Heinz inequality itself.

Theorem 1.3 [3]

Let $0\le p$, $1\le q$ and $0\le r$ with $p+r\le \left(1+r\right)q$. If $0\le B\le A$, then
${\left({A}^{\frac{r}{2}}{B}^{p}{A}^{\frac{r}{2}}\right)}^{\frac{1}{q}}\le {A}^{\frac{p+r}{q}}.$
(2)
The following result by Tanahashi is a full description of the best possibility of the range
$p+r\le \left(1+r\right)q\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}1\le q$

as far as all parameters are positive.

Theorem 1.4 [4]

Let p, q, r be positive real numbers. If $\left(1+r\right)q or $0, then there exist $2×2$ matrices A, B with $0 that do not satisfy the inequality
${\left({A}^{\frac{r}{2}}{B}^{p}{A}^{\frac{r}{2}}\right)}^{\frac{1}{q}}\le {A}^{\frac{p+r}{q}}.$

One notices the coincidence between the assumption on parameters in Theorem 1.1 and Theorem 1.3. As a matter of fact, the inequality (1) is a particular conclusion of the Furuta inequality. We should point out that Tanahashi’s argument in [4] is almost sufficient to deduce the former from the latter. In the next section, we will prove Theorem 1.1 using Theorem 1.3 and Tanahashi’s argument.

## 2 Proof of Theorem 1.1

As we mentioned above, our proof of Theorem 1.1 has a major part which is parallel to [4]. Our matrix A is a little different from that in [4], we use a variable y instead of ε and δ. It simplifies the argument to an extent, though the improvement is not essential.

Throughout this paper, we assume that $1 and $0. We will consider matrices
$A=\left(\begin{array}{cc}a& \sqrt{\left(a-1\right)y}\\ \sqrt{\left(a-1\right)y}& b+y\end{array}\right)$
and
$B=\left(\begin{array}{cc}1& 0\\ 0& b\end{array}\right).$

Then we have $0. The eigenvalues of A are $\frac{a+b+y±\sqrt{d}}{2}$, where $d={a}^{2}+{b}^{2}+{y}^{2}-2ab+2\left(a+b-2\right)y$.

Lemma 2.1 $0 and $a-b-y-\sqrt{d}\ne 0$.

Proof Obviously,
$\begin{array}{c}d={\left(a-b\right)}^{2}+y\left(y+2\left(a+b-2\right)\right)>0,\hfill \\ d={\left(a+b+y\right)}^{2}-4\left(ab+y\right)<{\left(a+b+y\right)}^{2}.\hfill \end{array}$

If $a-b-y-\sqrt{d}=0$, then we would have $a=1$ or $y=0$, which is contrary to the assumption. □

Let
$c=\frac{-2\sqrt{\left(a-1\right)y}}{a-b-y-\sqrt{d}}$
and
$U=\frac{1}{\sqrt{{c}^{2}+1}}\left(\begin{array}{cc}c& 1\\ 1& -c\end{array}\right).$
Then U is unitary and
${U}^{\ast }AU=\frac{1}{2}\left(\begin{array}{cc}{d}_{1}& 0\\ 0& {d}_{2}\end{array}\right),$
where
${d}_{1}=a+b+y+\sqrt{d},\phantom{\rule{2em}{0ex}}{d}_{2}=a+b+y-\sqrt{d}.$
By the assumption and Theorem 1.3, A and B satisfy the inequality (2). Then
${\left({U}^{\ast }{A}^{\frac{r}{2}}U{U}^{\ast }{B}^{p}U{U}^{\ast }{A}^{\frac{r}{2}}U\right)}^{\frac{1}{q}}\le {U}^{\ast }{A}^{\frac{p+r}{q}}U,$
hence we have
${\left\{\left(\begin{array}{cc}{d}_{1}^{\frac{r}{2}}& 0\\ 0& {d}_{2}^{\frac{r}{2}}\end{array}\right){U}^{\ast }\left(\begin{array}{cc}1& 0\\ 0& {b}^{p}\end{array}\right)U\left(\begin{array}{cc}{d}_{1}^{\frac{r}{2}}& 0\\ 0& {d}_{2}^{\frac{r}{2}}\end{array}\right)\right\}}^{\frac{1}{q}}\le {2}^{-\frac{p}{q}}\left(\begin{array}{cc}{d}_{1}^{\frac{p+r}{q}}& 0\\ 0& {d}_{2}^{\frac{p+r}{q}}\end{array}\right).$
(3)
Denote
$\left(\begin{array}{cc}{d}_{1}^{\frac{r}{2}}& 0\\ 0& {d}_{2}^{\frac{r}{2}}\end{array}\right){U}^{\ast }\left(\begin{array}{cc}1& 0\\ 0& {b}^{p}\end{array}\right)U\left(\begin{array}{cc}{d}_{1}^{\frac{r}{2}}& 0\\ 0& {d}_{2}^{\frac{r}{2}}\end{array}\right)=\frac{1}{{c}^{2}+1}\left(\begin{array}{cc}{A}_{1}& {A}_{3}\\ {A}_{3}& {A}_{2}\end{array}\right),$
where
$\begin{array}{c}{A}_{1}={d}_{1}^{r}\left({c}^{2}+{b}^{p}\right),\hfill \\ {A}_{2}={d}_{2}^{r}\left(1+{c}^{2}{b}^{p}\right),\hfill \\ {A}_{3}={d}_{1}^{\frac{r}{2}}{d}_{2}^{\frac{r}{2}}c\left(1-{b}^{p}\right)={\left({\left(a+b+y\right)}^{2}-d\right)}^{\frac{r}{2}}c\left(1-{b}^{p}\right)={\left(4ab+4y\right)}^{\frac{r}{2}}c\left(1-{b}^{p}\right).\hfill \end{array}$

Lemma 2.2 Let p, q, r be positive real numbers. Then ${A}_{2}<{A}_{1}$ and ${A}_{3}<0$.

Proof Since ${d}_{2}<{d}_{1}$ and $0, we have ${d}_{2}^{r}<{d}_{1}^{r}$. Moreover,
$\left({c}^{2}+{b}^{p}\right)-\left(1+{c}^{2}{b}^{p}\right)=\left({c}^{2}-1\right)\left(1-{b}^{p}\right),\phantom{\rule{1em}{0ex}}1-{b}^{p}<0$
and
${c}^{2}-1=-\frac{2{\left(a-b\right)}^{2}+2{y}^{2}+4\left(b-a\right)y+2\left(b-a+y\right)\sqrt{d}}{{\left(a-b-y-\sqrt{d}\right)}^{2}}<0,$

hence we have $1+{c}^{2}{b}^{p}<{c}^{2}+{b}^{p}$. Thus ${A}_{2}<{A}_{1}$.

It is obvious that $1-{b}^{p}<0$ and $0, and hence ${A}_{3}<0$. □

Let
$V=\frac{1}{\sqrt{{A}_{1}-{A}_{2}+2{\epsilon }_{1}}}\left(\begin{array}{cc}\sqrt{{A}_{1}-{A}_{2}+{\epsilon }_{1}}& -\sqrt{{\epsilon }_{1}}\\ -\sqrt{{\epsilon }_{1}}& -\sqrt{{A}_{1}-{A}_{2}+{\epsilon }_{1}}\end{array}\right),$
where
$2{\epsilon }_{1}=-{A}_{1}+{A}_{2}+\sqrt{{\left({A}_{1}-{A}_{2}\right)}^{2}+4{A}_{3}^{2}}.$
Then it is easy to see that ${A}_{3}=-\sqrt{\left({A}_{1}-{A}_{2}+{\epsilon }_{1}\right){\epsilon }_{1}}$, V is unitary and
${V}^{\ast }\left(\begin{array}{cc}{A}_{1}& {A}_{3}\\ {A}_{3}& {A}_{2}\end{array}\right)V=\left(\begin{array}{cc}{A}_{1}+{\epsilon }_{1}& 0\\ 0& {A}_{2}-{\epsilon }_{1}\end{array}\right).$

The following lemma is one of the most important points in Tanahashi’s argument. Although the substance is presented in the whole proof of [4], Theorem], we should restate and prove it in our context for the readers’ convenience.

Lemma 2.3
$\begin{array}{r}{\epsilon }_{1}\left\{\gamma {d}_{1}^{\frac{p+r}{q}}-{\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}}\right\}\left\{{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}-\gamma {d}_{2}^{\frac{p+r}{q}}\right\}\\ \phantom{\rule{1em}{0ex}}\le \left({A}_{1}-{A}_{2}+{\epsilon }_{1}\right)\left\{\gamma {d}_{1}^{\frac{p+r}{q}}-{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}\right\}\left\{\gamma {d}_{2}^{\frac{p+r}{q}}-{\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}}\right\},\end{array}$
(4)

where $\gamma ={\left(\frac{{c}^{2}+1}{{2}^{p}}\right)}^{\frac{1}{q}}$.

Proof The formula (3) implies
${\left({c}^{2}+1\right)}^{-\frac{1}{q}}V\left(\begin{array}{cc}{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}& 0\\ 0& {\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}}\end{array}\right){V}^{\ast }\le {2}^{-\frac{p}{q}}\left(\begin{array}{cc}{d}_{1}^{\frac{p+r}{q}}& 0\\ 0& {d}_{2}^{\frac{p+r}{q}}\end{array}\right).$
(5)
Write the left-hand matrix as
${\left({c}^{2}+1\right)}^{-\frac{1}{q}}{\left({A}_{1}-{A}_{2}+2{\epsilon }_{1}\right)}^{-1}\left(\begin{array}{cc}{B}_{1}& {B}_{3}\\ {B}_{3}& {B}_{2}\end{array}\right),$
where
$\begin{array}{c}{B}_{1}=\left({A}_{1}-{A}_{2}+{\epsilon }_{1}\right){\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}+{\epsilon }_{1}{\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}},\hfill \\ {B}_{2}={\epsilon }_{1}{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}+\left({A}_{1}-{A}_{2}+{\epsilon }_{1}\right){\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}},\hfill \\ {B}_{3}=-\sqrt{{A}_{1}-{A}_{2}+{\epsilon }_{1}}\sqrt{{\epsilon }_{1}}\left\{{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}-{\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}}\right\}.\hfill \end{array}$
Then, by the formula (5), we have
$0\le \left(\begin{array}{cc}\gamma \left({A}_{1}-{A}_{2}+2{\epsilon }_{1}\right){d}_{1}^{\frac{p+r}{q}}-{B}_{1}& -{B}_{3}\\ -{B}_{3}& \gamma \left({A}_{1}-{A}_{2}+2{\epsilon }_{1}\right){d}_{2}^{\frac{p+r}{q}}-{B}_{2}\end{array}\right).$
So, its determinant is also non-negative. We expand it to obtain
$\begin{array}{rcl}0& \le & {\gamma }^{2}{\left({A}_{1}-{A}_{2}+2{\epsilon }_{1}\right)}^{2}{d}_{1}^{\frac{p+r}{q}}{d}_{2}^{\frac{p+r}{q}}-\gamma \left({A}_{1}-{A}_{2}+2{\epsilon }_{1}\right){d}_{1}^{\frac{p+r}{q}}{B}_{2}\\ -\gamma \left({A}_{1}-{A}_{2}+2{\epsilon }_{1}\right){d}_{2}^{\frac{p+r}{q}}{B}_{1}+{B}_{1}{B}_{2}-{B}_{3}^{2}.\end{array}$
(6)
Now,
$\begin{array}{r}{B}_{1}{B}_{2}-{B}_{3}^{2}\\ \phantom{\rule{1em}{0ex}}=\left\{\left({A}_{1}-{A}_{2}+{\epsilon }_{1}\right){\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}+{\epsilon }_{1}{\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}}\right\}\left\{{\epsilon }_{1}{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}+\left({A}_{1}-{A}_{2}+{\epsilon }_{1}\right){\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}}\right\}\\ \phantom{\rule{2em}{0ex}}-\left({A}_{1}-{A}_{2}+{\epsilon }_{1}\right){\epsilon }_{1}{\left\{{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}-{\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}}\right\}}^{2}\\ \phantom{\rule{1em}{0ex}}={\left({A}_{1}-{A}_{2}+2{\epsilon }_{1}\right)}^{2}{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}{\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}}.\end{array}$
Hence, the formula (6) implies
$\begin{array}{rl}0\le & \left({A}_{1}-{A}_{2}+2{\epsilon }_{1}\right)\left\{{\gamma }^{2}\left({A}_{1}-{A}_{2}+2{\epsilon }_{1}\right){d}_{1}^{\frac{p+r}{q}}{d}_{2}^{\frac{p+r}{q}}-\gamma {d}_{1}^{\frac{p+r}{q}}{B}_{2}-\gamma {d}_{2}^{\frac{p+r}{q}}{B}_{1}\right\}\\ +{\left({A}_{1}-{A}_{2}+2{\epsilon }_{1}\right)}^{2}{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}{\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}}.\end{array}$
Cancel the common positive factor ${A}_{1}-{A}_{2}+2{\epsilon }_{1}$ and substitute the definitions for ${B}_{1}$ and ${B}_{2}$. Then a simple calculation shows that
$\begin{array}{r}-{\epsilon }_{1}\left\{{\gamma }^{2}{d}_{1}^{\frac{p+r}{q}}{d}_{2}^{\frac{p+r}{q}}-\gamma {d}_{1}^{\frac{p+r}{q}}{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}-\gamma {d}_{2}^{\frac{p+r}{q}}{\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}}+{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}{\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}}\right\}\\ \phantom{\rule{1em}{0ex}}\le \left({A}_{1}-{A}_{2}+{\epsilon }_{1}\right)\\ \phantom{\rule{2em}{0ex}}\cdot \left\{{\gamma }^{2}{d}_{1}^{\frac{p+r}{q}}{d}_{2}^{\frac{p+r}{q}}-\gamma {d}_{1}^{\frac{p+r}{q}}{\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}}-\gamma {d}_{2}^{\frac{p+r}{q}}{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}+{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}{\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}}\right\}.\end{array}$
By factorizing, we have
$\begin{array}{r}-{\epsilon }_{1}\left\{\gamma {d}_{1}^{\frac{p+r}{q}}-{\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}}\right\}\left\{\gamma {d}_{2}^{\frac{p+r}{q}}-{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}\right\}\\ \phantom{\rule{1em}{0ex}}\le \left({A}_{1}-{A}_{2}+{\epsilon }_{1}\right)\left\{\gamma {d}_{1}^{\frac{p+r}{q}}-{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}\right\}\left\{\gamma {d}_{2}^{\frac{p+r}{q}}-{\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}}\right\}.\end{array}$

This completes the proof of Lemma 2.3. □

Now, we estimate each term of the inequality (4) with respect to $y\to +0$. A key point in making use of the inequality (4) is that both estimations of the factor ${\epsilon }_{1}$ on the left-hand side and the factor $\gamma {d}_{1}^{\frac{p+r}{q}}-{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}$ on the right-hand side contain a common subfactor y. After the cancellation of this y, we will derive the desired functional inequality by letting $y\to +0$, $a\to 1+0$ and applying l’Hopital’s rule. Terms in other factors can be roughly estimated.

In the following, o means $o\left(y\right)$, that is
$\frac{o}{y}\to 0\phantom{\rule{1em}{0ex}}\left(y\to +0\right),$

and $o\left(1\right)$ denotes a term such that $o\left(1\right)\to 0$ ($y\to +0$).

One can establish the following formulae:
$\begin{array}{c}\sqrt{d}=\left(b-a\right)\left\{1+\frac{a+b-2}{{\left(b-a\right)}^{2}}y+o\left(y\right)\right\},\hfill \\ {d}_{1}^{\frac{p+r}{q}}={\left(2b\right)}^{\frac{p+r}{q}}\left\{1+\frac{p+r}{q}\cdot \frac{b-1}{b\left(b-a\right)}y+o\left(y\right)\right\},\hfill \\ {d}_{2}^{\frac{p+r}{q}}={\left(2a\right)}^{\frac{p+r}{q}}\left\{1+\frac{p+r}{q}\cdot \frac{-a+1}{a\left(b-a\right)}y+o\left(y\right)\right\},\hfill \\ c=\frac{-2\sqrt{\left(a-1\right)y}}{a-b-y-\left(b-a+\frac{a+b-2}{b-a}y+o\left(y\right)\right)}=\sqrt{y}\cdot \frac{\sqrt{a-1}}{b-a}\left\{1-\frac{b-1}{{\left(b-a\right)}^{2}}y+o\left(y\right)\right\},\hfill \\ {c}^{2}+1=1+\frac{a-1}{{\left(b-a\right)}^{2}}y+o\left(y\right),\hfill \\ \begin{array}{r}{\left({c}^{2}+1\right)}^{\frac{1}{q}}{d}_{1}^{\frac{p+r}{q}}\\ \phantom{\rule{1em}{0ex}}=\left\{1+\frac{a-1}{q{\left(b-a\right)}^{2}}y+o\left(y\right)\right\}{\left(2b\right)}^{\frac{p+r}{q}}\left\{1+\frac{p+r}{q}\cdot \frac{b-1}{b\left(b-a\right)}y+o\left(y\right)\right\}\\ \phantom{\rule{1em}{0ex}}={\left(2b\right)}^{\frac{p+r}{q}}\left\{1+\frac{1}{qb{\left(b-a\right)}^{2}}\left(\left(a-1\right)b+\left(p+r\right)\left(b-1\right)\left(b-a\right)\right)y+o\left(y\right)\right\},\end{array}\hfill \\ {\left({c}^{2}+1\right)}^{\frac{1}{q}}{d}_{2}^{\frac{p+r}{q}}={\left(2a\right)}^{\frac{p+r}{q}}\left(1+o\left(1\right)\right),\hfill \\ \begin{array}{rl}{A}_{1}& ={\left(2b\right)}^{r}\left\{1+\frac{r\left(b-1\right)}{b\left(b-a\right)}y+o\left(y\right)\right\}\left\{{b}^{p}+\frac{a-1}{{\left(b-a\right)}^{2}}y+o\left(y\right)\right\}\\ ={2}^{r}{b}^{p+r}\left\{1+\frac{1}{b{\left(b-a\right)}^{2}}\left(r\left(b-1\right)\left(b-a\right)+{b}^{1-p}\left(a-1\right)\right)y+o\left(y\right)\right\},\end{array}\hfill \\ {A}_{2}={\left(2a\right)}^{r}\left(1+o\left(1\right)\right),\hfill \\ {A}_{3}^{2}={\left(4ab+4y\right)}^{r}y\frac{a-1}{{\left(b-a\right)}^{2}}\left(1+o\left(1\right)\right){\left(1-{b}^{p}\right)}^{2}=y{4}^{r}{a}^{r}{b}^{r}\frac{a-1}{{\left(b-a\right)}^{2}}{\left(1-{b}^{p}\right)}^{2}\left(1+o\left(1\right)\right),\hfill \\ \begin{array}{rl}{\epsilon }_{1}& =\frac{1}{2}\left({A}_{1}-{A}_{2}\right)\left(-1+\sqrt{1+\frac{4{A}_{3}^{2}}{{\left({A}_{1}-{A}_{2}\right)}^{2}}}\right)=\frac{{A}_{3}^{2}}{{A}_{1}-{A}_{2}}+o\\ =\frac{y{4}^{r}{a}^{r}{b}^{r}\left(a-1\right){\left(b-a\right)}^{-2}{\left(1-{b}^{p}\right)}^{2}\left(1+o\left(1\right)\right)}{{2}^{r}{b}^{p+r}\left(1+o\left(1\right)\right)-{\left(2a\right)}^{r}\left(1+o\left(1\right)\right)}+o\\ =\frac{y{2}^{r}{a}^{r}{b}^{r}\left(a-1\right){\left(1-{b}^{p}\right)}^{2}}{{\left(b-a\right)}^{2}\left({b}^{p+r}-{a}^{r}\right)}\left(1+o\left(1\right)\right),\end{array}\hfill \\ \begin{array}{r}{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}\\ \phantom{\rule{1em}{0ex}}=\left({2}^{r}{b}^{p+r}\left\{1+\frac{1}{b{\left(b-a\right)}^{2}}\left(r\left(b-1\right)\left(b-a\right)+{b}^{1-p}\left(a-1\right)\right)y+o\left(y\right)\right\}\\ \phantom{\rule{2em}{0ex}}+\frac{y{2}^{r}{a}^{r}{b}^{r}\left(a-1\right){\left(1-{b}^{p}\right)}^{2}}{{\left(b-a\right)}^{2}\left({b}^{p+r}-{a}^{r}\right)}\left(1+o\left(1\right)\right)\right){\phantom{\rule{0.2em}{0ex}}}^{\frac{1}{q}}\\ \phantom{\rule{1em}{0ex}}={2}^{\frac{r}{q}}{b}^{\frac{p+r}{q}}\left\{1+\frac{1}{qb{\left(b-a\right)}^{2}}\left(r\left(b-1\right)\left(b-a\right)+{b}^{1-p}\left(a-1\right)+\frac{{a}^{r}{b}^{1-p}\left(a-1\right){\left(1-{b}^{p}\right)}^{2}}{{b}^{p+r}-{a}^{r}}\right)y\\ \phantom{\rule{2em}{0ex}}+o\left(y\right)\right\},\end{array}\hfill \\ {\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}}={2}^{\frac{r}{q}}{a}^{\frac{r}{q}}\left(1+o\left(1\right)\right),\hfill \\ {A}_{1}-{A}_{2}+{\epsilon }_{1}={2}^{r}\left({b}^{p+r}-{a}^{r}\right)\left(1+o\left(1\right)\right),\hfill \\ \gamma {d}_{1}^{\frac{p+r}{q}}-{\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}}={2}^{\frac{r}{q}}\left({b}^{\frac{p+r}{q}}-{a}^{\frac{r}{q}}\right)\left(1+o\left(1\right)\right),\hfill \\ \gamma {d}_{2}^{\frac{p+r}{q}}-{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}={2}^{\frac{r}{q}}\left({a}^{\frac{p+r}{q}}-{b}^{\frac{p+r}{q}}\right)\left(1+o\left(1\right)\right),\hfill \\ \gamma {d}_{2}^{\frac{p+r}{q}}-{\left({A}_{2}-{\epsilon }_{1}\right)}^{\frac{1}{q}}={2}^{\frac{r}{q}}\left({a}^{\frac{p+r}{q}}-{a}^{\frac{r}{q}}\right)\left(1+o\left(1\right)\right).\hfill \end{array}$
Now, we have the estimation of the most delicate factor in the formula (4), whose constant term is canceled by subtraction.
$\begin{array}{r}\gamma {d}_{1}^{\frac{p+r}{q}}-{\left({A}_{1}+{\epsilon }_{1}\right)}^{\frac{1}{q}}\\ \phantom{\rule{1em}{0ex}}={2}^{-\frac{p}{q}}\cdot {\left(2b\right)}^{\frac{p+r}{q}}\left\{1+\frac{1}{qb{\left(b-a\right)}^{2}}\left(\left(a-1\right)b+\left(p+r\right)\left(b-1\right)\left(b-a\right)\right)y+o\left(y\right)\right\}\\ \phantom{\rule{2em}{0ex}}-{2}^{\frac{r}{q}}{b}^{\frac{p+r}{q}}\left\{1+\frac{1}{qb{\left(b-a\right)}^{2}}\left(r\left(b-1\right)\left(b-a\right)+{b}^{1-p}\left(a-1\right)+\frac{{a}^{r}{b}^{1-p}\left(a-1\right){\left(1-{b}^{p}\right)}^{2}}{{b}^{p+r}-{a}^{r}}\right)y\\ \phantom{\rule{2em}{0ex}}+o\left(y\right)\right\}\\ \phantom{\rule{1em}{0ex}}={2}^{\frac{r}{q}}\frac{{b}^{\frac{p+r}{q}-1}}{q{\left(b-a\right)}^{2}}\left\{\left(a-1\right)b+p\left(b-1\right)\left(b-a\right)-{b}^{1-p}\left(a-1\right)-\frac{{a}^{r}{b}^{1-p}\left(a-1\right){\left(1-{b}^{p}\right)}^{2}}{{b}^{p+r}-{a}^{r}}\right\}y\\ \phantom{\rule{2em}{0ex}}\cdot \left(1+o\left(1\right)\right)\end{array}$
Substitute these estimations for the inequality (4), cancel the positive factor y, and let $y\to +0$, then we have
$\begin{array}{r}\frac{{2}^{r}{a}^{r}{b}^{r}\left(a-1\right){\left(1-{b}^{p}\right)}^{2}}{{\left(b-a\right)}^{2}\left({b}^{p+r}-{a}^{r}\right)}\cdot {2}^{\frac{r}{q}}\left({b}^{\frac{p+r}{q}}-{a}^{\frac{r}{q}}\right)\cdot {2}^{\frac{r}{q}}\left({b}^{\frac{p+r}{q}}-{a}^{\frac{p+r}{q}}\right)\\ \phantom{\rule{1em}{0ex}}\le {2}^{r}\left({b}^{p+r}-{a}^{r}\right)\\ \phantom{\rule{2em}{0ex}}\cdot {2}^{\frac{r}{q}}\frac{{b}^{\frac{p+r}{q}-1}}{q{\left(b-a\right)}^{2}}\left\{\left(a-1\right)b+p\left(b-1\right)\left(b-a\right)-{b}^{1-p}\left(a-1\right)-\frac{{a}^{r}{b}^{1-p}\left(a-1\right){\left(1-{b}^{p}\right)}^{2}}{{b}^{p+r}-{a}^{r}}\right\}\\ \phantom{\rule{2em}{0ex}}\cdot {2}^{\frac{r}{q}}\left({a}^{\frac{p+r}{q}}-{a}^{\frac{r}{q}}\right),\end{array}$
and hence
$\begin{array}{r}{a}^{r}{b}^{r}{\left(1-{b}^{p}\right)}^{2}\cdot \left({b}^{\frac{p+r}{q}}-{a}^{\frac{r}{q}}\right)\cdot \left({b}^{\frac{p+r}{q}}-{a}^{\frac{p+r}{q}}\right)\\ \phantom{\rule{1em}{0ex}}\le {\left({b}^{p+r}-{a}^{r}\right)}^{2}\\ \phantom{\rule{2em}{0ex}}\cdot \frac{{b}^{\frac{p+r}{q}-1}}{q}\left\{\left(a-1\right)b+p\left(b-1\right)\left(b-a\right)-{b}^{1-p}\left(a-1\right)-\frac{{a}^{r}{b}^{1-p}\left(a-1\right){\left(1-{b}^{p}\right)}^{2}}{{b}^{p+r}-{a}^{r}}\right\}\\ \phantom{\rule{2em}{0ex}}\cdot \frac{{a}^{\frac{p+r}{q}}-{a}^{\frac{r}{q}}}{a-1}.\end{array}$
Letting $a\to 1+0$ and applying l’Hopital’s rule, we have
${b}^{r}{\left(1-{b}^{p}\right)}^{2}{\left({b}^{\frac{p+r}{q}}-1\right)}^{2}\le {\left({b}^{p+r}-1\right)}^{2}{b}^{\frac{p+r}{q}-1}\frac{{p}^{2}}{{q}^{2}}{\left(b-1\right)}^{2}.$
This implies that, for arbitrary $1,
${b}^{\frac{1+r-\frac{p+r}{q}}{2}}\left({b}^{p}-1\right)\left({b}^{\frac{p+r}{q}}-1\right)\le \frac{p}{q}\left({b}^{p+r}-1\right)\left(b-1\right).$
(7)

For arbitrary $0, substitute $\frac{1}{x}$ for b in (7) and multiply by x, ${x}^{p}$, ${x}^{p+r}$, ${x}^{\frac{p+r}{q}}$ both sides. It is easy to see that x itself satisfies (7). This completes the proof of Theorem 1.1.

## Declarations

### Acknowledgements

The author was supported in part by Grants-in-Aid for Scientific Research, Japan Society for the Promotion of Science.

## Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Niigata University, Niigata 950-2181, Japan

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