Let $p\in \mathbb R_{\geq2}$, $x>0$, prove or disprove $$\left(x+\dfrac{1}{x}\right)^p-x^p-\dfrac{1}{x^p}\ge 2^p-2$$

I can prove this for positive integers $p$ because we can use $$\left(x+\dfrac{1}{x}\right)^p-x^p-\dfrac{1}{x^p}=\dfrac{1}{2}\sum_{i=1}^{p-1}\binom{p}{i}(x^{2i-p}+x^{p-2i})\ge\sum_{i=1}^{p-1} \binom{p}{i}=2^p-2$$

But is this true for all $p\in \mathbb{R}_{\geq 2}$?

  • $\begingroup$ I think you are right.You have done absolutely right. $\endgroup$ – Sufaid Saleel Aug 7 '17 at 15:24
  • $\begingroup$ @SufaidSaleel,Thanks,How to prove it? $\endgroup$ – wightahtl Aug 7 '17 at 15:25
  • $\begingroup$ You are given that $p\geq 2$, which is certainly positive. $\endgroup$ – amWhy Aug 7 '17 at 16:58
  • $\begingroup$ @amWhy "$p$ positive" is not sufficient. The inequality is trivially true for $p=1$ and $p=2$, but not for $1<p<2$. ;-) $\endgroup$ – user436658 Aug 7 '17 at 17:22
  • $\begingroup$ Look at the first sentence in the post DrMV: "Let $p\ge 2,p\in \mathbb R, \;x>0$".... $\endgroup$ – amWhy Aug 7 '17 at 19:09

I will reduce the question to some known results. For positive numbers $a,b$ we define the geometric mean $G(a,b)$ and the power mean of order $s$ denoted by $M_s(a,b)$ as folows $$G(a,b)=\sqrt{ab},\qquad M_s(a,b)=\left(\frac{a^s+b^s}{2}\right)^{1/s}$$ Now, with $a=x$, $b=1/x$, and writing $M_s$ and $G$ instead of $M_s(a,b)$ and $G(a,b)$ for simplicity, the proposed inequality takes the form $$\frac{M_1^p-G^p}{M_p^p-G^p}\ge 2^{1-p}$$ So, this is a particular case of the more general inequality

$$\frac{M_s^p-G^p}{M_t^p-G^p}\ge 2^{p/t-p/s}$$ Which is valid for $0<s<t$, $p\ge \max(2s,2(s+t)/3)$. A detailed proof and more results on this topic can be found in the paper of Omran Kouba entitled: "Bounds for the ratios of differences of power means in two arguments"

This paper can be found here or here.

  • 1
    $\begingroup$ Very nice solution!! $\endgroup$ – Andreas Aug 11 '17 at 8:38

Starting with $L^p$ norm inequality $$ (x^p +x^{-p})^{1/p} \leq x +x^{-1},\;\; p \geq 2 \geq 1,\;\; \infty > x > 0, $$ Or, (since both sides are positive) $$ x^p +x^{-p} \leq (x +x^{-1})^p $$ Consider, $$ L(x) :=(x +x^{-1})^p -(x^p +x^{-p}) \geq 0, $$ Particularly, when $x=1$, $$ 2^p -2 \geq 0.\quad \quad \ldots (\heartsuit) $$ Differentiate $L(x)$, with straightforward arrangement, $$ \frac{dL(x)}{dx} = \ldots =p x^{-p-1} \big[ (x^2+1)^{p-1} (x^2-1) -(x^{2p} -1) \big]. $$ We hope that (within considered range) $$ \frac{dL(x)}{dx} \geq 0 \quad \quad \ldots (\clubsuit) $$ We first focus on the case $x>1$. Then it amounts to show $$ (x^2+1)^{p-1} (x^2-1) \geq x^{2p} -1,\quad \quad \ldots (\spadesuit') $$ For, if so, by $(\heartsuit)$, $(\clubsuit)$, and $(\spadesuit')$ $$ (x +x^{-1})^p -(x^p +x^{-p}) \geq 2^p -2 $$ As desired.

Come back to $(\spadesuit')$. With $$ y :=\frac{x^2-1}{2(x^2+1)} =\frac{1}{2} -\frac{1}{x^2+1} $$ implying $$ 0 <y <\frac{1}{2} $$ (My motivation of substitution came from the desire that the whole inequality shall be written in a form with "homogeneous dimensions", and range of variables shall lie in a finite interval.)

After some completely straightforward but tedious manipulation, $(\spadesuit')$ becomes $$ 1 \geq \frac{1}{2y} \left[ \left( \frac{1}{2} +y \right)^p -\left( \frac{1}{2} -y \right)^p \right]. \quad \quad \ldots (\spadesuit) $$ Now, this is interpreted with the fact that, in the graph of $x^p$, a secant line lying in the very middle of $[0,1]$ has slope less than 1, which is that of the line from 0 to 1.

To show this, find $\eta_1,\eta_2$, according to mean value theorem, $$ p \eta_1^{p-1} y =\left( \frac{1}{2} +y \right)^p -\left( \frac{1}{2} \right)^p, \\ p \eta_2^{p-1} y =\left( \frac{1}{2} \right)^p -\left( \frac{1}{2} -y \right)^p $$ Thus rhs of $(\spadesuit)$ becomes (to stress dependence on $y$) $$ \frac{p}{2} (\eta_1(y)^{p-1} -\eta_2(y)^{p-1}) $$ Of course, $\eta_1(y) >\eta_2(y)$. And observe that $\eta_1(y)$ is increasing with $y$. Indeed, the secant line with greater $y$ must correspond to a greater $\eta_1^{p-1}$, thus a greater $\eta_1$. Similarly, $\eta_1(y)$ is decreasing with $y$. So, rhs of $(\spadesuit)$, seen as function of $y$, is increasing too. The maximum value is achieved when $y=1/2$, which gives 1.

The case $x=1$ is trivial, and the case $0<x<1$ can be obtained by $x \mapsto x^{-1}$ and what we just showed.


Here is an answer which again makes use of the derivative, but in a different fashion than in Aminopterin's answer. Let

$$L(x) = \left(x+\dfrac{1}{x}\right)^p-x^p-\dfrac{1}{x^p} - 2^p+2$$

and we want to establish that $L(x) \ge 0$. Note that $L(x=1) = 0$.

We have (see Aminopterin's answer) $$ L'(x) = \frac{dL(x)}{dx} = p x^{-p-1} \big[ (x^2+1)^{p-1} (x^2-1) -(x^{2p} -1) \big] $$

Now we can show that $L(x=1) = 0$ is indeed the only condition with equality in the OP's problem if we can establish that $L'(x) > 0$ for $x > 1$ and $L'(x) < 0$ for $0<x < 1$. To this end, it suffices to set $x^2 = y$ and consider

$$ f(y) = (y+1)^{p-1} (y-1) - y^{p}+1 $$

We need to establish that $f(y) > 0$ for $y > 1$ and $f(y) < 0$ for $0<y < 1$. Now this has been established here, which completes the proof. $\qquad \Box$


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