TITLE: Prove or Disprove $(x+\frac{1}{x})^p-x^p-\frac{1}{x^p}\ge 2^p-2$
QUESTION [6 upvotes]: 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 $p\in \mathbb{R}_{\geq 2}$?

REPLY [3 votes]: 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.