For $p \in [1,2]$, how does one show $\sup\limits_{x\in\mathbb{R}}\frac{|1+x|^p-1-px}{|x|^p}\leq 2^{2-p}?$ The following question arose from Korolëiìuk's Theory of U-statistics:

For $p\in[1,2]$, how can we show that
  $$\sup_{x\in\mathbb{R}}\frac{|1+x|^p-1-px}{|x|^p}\leq 2^{2-p}?$$


My attempts: I tried some rearrangements such as $|1+x|^p\leq 1+px+2^{2-p}|x|^p$ and see some relation by manipulations but stuck at some point.
 A: for $|1+x|^p\leq 1+px+2^{2-p}|x|^p$, it is trivial the "=" holds when $x=0$ or $p=2$, exclude these cases, we can discuss 3 different cases:
1)$x>0$
$f(x)= 1+px+2^{2-p}x^p-(1+x)^p,f'(x)=p+p(2^{2-p}x^{p-1}-(1+x)^{p-1}),f"(x)=p(p-1)(2^{2-p}x^{p-2}-(1+x)^{p-2})$
$p(p-1)\ge0, f"(x)\ge0 \iff 2^{2-p}x^{p-2}-(1+x)^{p-2}\ge 0 \iff 2^{2-p}>(\dfrac{x}{1+x})^{2-p}$<1>
$2-p>0,$  so <1> is true $\implies f'(x)>f'(0)=2^{2-p}p>0 \implies f(x)>f(0)=0$
2)$-1<x<0$
let $t=1+x, \implies 0<t<1$ ,the inequality becomes:
$t^p \le 1 +pt-p+2^{2-p}(1-t)^p ,g(t)=1 +pt-p+2^{2-p}(1-t)^p-t^p,g'(t)=p(1-2^{2-p}(1-t)^{p-1}-t^{p-1}),g"(t)=p(p-1)(2^{2-p}(1-t)^{p-2}-t^{p-2}) $
$g(t)"=0 \to t= \dfrac{1}{3}$ to verify $2-\dfrac{1-t}{t}$,we know $g"(\dfrac{1}{3})$ is min point. so max point of $g'(t)$  is on $t=0$ or $t=1$
$g'(0)=-2^{2-p},g'(1)=0 \implies g'(t)_{max}=0 \implies g'(t)<0 \implies g(t)_{min}=g(1)=0 \implies g(t)>0$
3)$x \le -1$
$u=-x\ge 1 , h(u)=1-pu+2^{2-p}u^p-(u-1)^p,h'(u)=p(-1+2^{2-p}(u)^{p-1}-(u-1)^{p-1}),\\h"(u)=p(p-1)(2^{2-p}(u)^{p-2}-(u-1)^{p-2}) = 0 \implies u = 2 $, 
it is easy to verify $h'(2)=0$ is min point $\implies h'(u)>0 \implies h(u) \ge h(1)=2^{2-p}+1-p>0$
QED.
