Prove or Disprove $(x+\frac{1}{x})^p-x^p-\frac{1}{x^p}\ge 2^p-2$ 
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 $p\in \mathbb{R}_{\geq 2}$?
 A: 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.
A: 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.
A: 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. $\qquad \Box$ 
