Inequality with absolute value [Zorich's book] 
Show that $|1+x|^p\geq 1+px+c_p\varphi_p(x),$ where $c_p$ is a
constant depending only on $p$, $$\varphi_p(x)= \begin{cases} |x|^2, &
 \text{if }|x|\leq 1, \\ |x|^p, & \text{if }|x|>1, \end{cases}$$ if
$1<p\leq 2$, and $\varphi_p(x)=|x|^p$ on $\mathbb{R}$ if $2<p$.

This problem is from Zorich's Mathematical Analysis 1 and comes after the chapter "The study of functions using the methods of differential calculus". I have tried some approaches for this problem but I failed. So I'd be thankful to see the complete solution.
 A: So firstly, we give a base result without demonstrating it, that is:

Lemma:
$$|1+x|^p >1+px$$ for all $x \ne 0$ and $p>1$

Based on this result, I will divide and conquer the inequality in many cases.

Case 1: When $|x|<1$ and $1<p\le 2$

By Cauchy's MVT, there are $x_1,x_2 \in (-1,1)$ such that:
$$\frac{ (1+x)^p-1-px}{x^2} =\frac{p(1+x_1)^{p-1}-p}{2x_1}=\frac{1}{2}p(p-1)(1+x_2)^{p-2} \ge \frac{1}{2}p(p-1)2^{p-2}$$
So
$$(1+x)^p-1-px \ge \frac{1}{2}p(p-1)2^{p-2} x^2$$

Case 2: When $p>2$
Let
$$g(x)=\frac{|1+x|^p-1-px}{|x|^p}$$
Clearly, $g$ is continuous and positive on $(-\infty,0)\cup(0,\infty]$. Besides,
$$\lim_{x \rightarrow 0} \frac{(1+x)^p-1-px}{|x|^p}=\lim_{x \rightarrow 0} \frac{\frac{(1+x)^p-1-px}{x^2}}{|x|^{p-2}} =\lim_{x \rightarrow 0} \frac{\frac{1}{2}p(p-1)}{0^+}= +\infty $$
and
$$\lim_{x \rightarrow -\infty} \frac{|1+x|^p-1-px}{|x|^p}=\lim_{x \rightarrow +\infty} \frac{|1+x|^p-1-px}{|x|^p}=1$$
Hence $g$ achieves a positive minimum on $\mathbb{R}$, thus there is a constant $c_p>0$ such that:
$$(1+x)^p-1-px \ge c_p|x|^p$$
for all $x$

Case 3: When $|x| \ge 1$ and $1 \le p \le 2$
Then again, $g$ is continuous and positive on $(-\infty,-1]\cup[1,+\infty)$, and
$$\lim_{x \rightarrow -\infty} \frac{|1+x|^p-1-px}{|x|^p}=\lim_{x \rightarrow +\infty} \frac{|1+x|^p-1-px}{|x|^p}=1$$
Hence $g$ achieves a positive minimum on $\mathbb{R}$, thus there is a constant $c_p>0$ such that:
$$(1+x)^p-1-px \ge c_p|x|^p$$
for all $|x| \ge 1$ $\square$

Proof of lemma
If $x <-1$, the inequality is obvious.
Now if $x \ge -1$. Let's analyze the differentiable function $h(x)= (1+x)^p-1-px$ on $[-1,+\infty)$, we see that $h'(x)=p\left[ (1+x)^{p-1}-1 \right]$.

Clearly $h'(x)< 0 $ on $[-1,0)$ and $h'(x)>0$ on $(0, +\infty)$ hence
$$0= h(0)<h(x) = (1+x)^p-1-px$$
for all $x \ne 0$
$\square$
A: There may be a clever way to do it. Here I only do some differential analysis to understand the behavior when $|x|$ is large enough.
For $p>2$  consider the function
$$\phi_p(x)=\frac{|1+x|^p-px-1}{|x|^p}$$
For $x>0$, we have that as $x\rightarrow0+$
$$
\phi_p(x)\sim \frac{p(1+x)^{p-1}-p}{px^{p-1}}\sim\frac{(x+1)^{p-2}}{x^{p-2}}\xrightarrow{x\rightarrow0+}\infty$$
and as $x\rightarrow\infty$
$$\lim_{x\rightarrow\infty}\phi_p(x)=\lim_{x\rightarrow\infty}\Big(\Big(1+\frac1x\Big)^p-px^{1-p}-x^{-p}\Big)=1$$
Differentiation gives
$$\begin{align}\phi'_p(x)&=\frac{x^p(p(1+x)^{p-1}-p)-px^{p-1}((1+x)^p-px-1)}{x^{2p}}\\
&=\frac{px(1+x)^{p-1} -xp-p(1+x)^p +p^2x+p}{x^{p-1}}\\
&=\frac{-p(1+x)^{p-1}+px(p-1)+p}{x^{p-1}}=-p\frac{(1+x)^{p-1}-(p-1)x -1}{x^{p-1}}
\end{align}$$
Since $(1+x)^q\geq 1+qx$ for $x>0$ and $q>1$ with equality only if $x=0$ (this can be obtained directly by noticing that $x\mapsto (1+x)^q-qx$ is increasing),  $\phi'_p(x)<0$, and so $\phi_p$ decreases.
Thus for some $a$ large enough  we have $\phi_p(x)>1$
for all $x>0$.
For $x\leq-1$,
$$
\phi_p(x)=\frac{(-x-1)^p-px-1}{(-x)^p}
$$
Let $u=-x$, so that $u\geq1$, and define
$$\psi_p(u)=\frac{(u-1)^p+pu-1}{u^p}$$
Clearly $\psi_p(u)>0$, $\psi_p(1)=p-1>0$, and  $$\lim_{u\rightarrow\infty}\psi_p(u)=\lim_{u\rightarrow\infty}\Big(1-\frac{1}{u}\Big)^p+pu^{1-p}-u^{-p}=1$$
Differentiation we get
$$\begin{align}
\psi'_p(u)&=\frac{u^p(p(u-1)^{p-1}+p)-pu^{p-1}((u-1)^p+pu-1)}{u^{2p}}\\
&=-p\frac{(u-1)^{p-1}+u(p-1)-1}{u^{p-1}}
\end{align}$$
Since $\phi'_p(1)-p(2-p)<0$ and $\psi_p'(u)>0$ for all $p$ large enough, $\psi_p$ attains a minimum value at some point $u^*>1$, after which $\phi_p(u)\nearrow1$. This means that
$$ 0<\phi_p(-u_*)\leq \phi_p(x)<1$$
for all $x\leq 1$.
It remains to see what happens for in $[-1,0)$.
A $x\rightarrow0-$
$$\phi_p(x)\sim -\frac{p(1+x)^{p-1}-p}{p(-x)^{p-1}}\sim\frac{(x+1)^{p-2}}{(-x)^{p-2}}\xrightarrow{x\rightarrow0+}\infty$$
Since
$$\begin{align}\phi'_p(x)&=\frac{(-x)^p(p(1+x)^{p-1}-p)+p(-x)^{p-1}((1+x)^p-px-1)}{(-x)^{2p}}\\
&=\frac{-px(1+x)^{p-1} +xp+p(1+x)^p -p^2x-p}{(-x)^{p-1}}\\
&=\frac{p(1+x)^{p-1}-px(p-1)-p}{(-x)^{p-1}}=p\frac{(1+x)^{p-1}-(p-1)x -1}{(-x)^{p-1}}
\end{align}$$
we have that $\phi'_p(x)>0$ (for  $(1+x)^q>qx+1$ for all $q>1$ and $0<x<1$ as we can see directly by showing that $x\mapsto (1+x)^q-qx$ is decreasing in $(-1,0)$. Consequently, $\phi_p(x)$ increases in $(-1,0)$ and so, $\phi_p(x)\geq \phi_p(-1)=p-1>1>\phi_p(-u^*)>0$
Putting things together, we obtain that
$\phi_p(x)\geq \phi_p(-u^*)$ and the conclusion follows for the case $p>2$.
The case $1<p\leq 2$ can be handle similarly.
A: Addendum to Nguyen's answer: I think that the proof of the case $p\in (1,2]$ can be done without Cauchy's MVT. Here is what I have done (please correct me I am wrong).
Consider the function $g:(-\infty,-1]\cup [1,+\infty)\to \mathbb{R}$ defined by $g(x)=\dfrac{|1+x|^p-1-px}{|x|^p}.$ Clearly $g(x)$ is continuous and positive on $(-\infty,-1]\cup [1,+\infty)$.
Since $\lim \limits_{x\to +\infty}g(x)=\lim \limits_{x\to -\infty}g(x)=1$ then $\exists \delta>1$ such that if $|x|\geq \delta$ then $\frac{1}{2}<g(x)<\frac{3}{2}$.
On $[-\delta, -1]$ the function $g(x)$ attains it minimum value $f(\xi_1)>0$ (by Weierstrass theorem) and also on the $[1,\delta]$, denote it by $f(\xi_2)>0$. Hence $g(x)\geq \min\{g(\xi_1),g(\xi_2),\frac{1}{2}\}>0$ for $|x|\geq 1$.
Hence $\exists a_p>0$ such that $g(x)\geq a_p$ for $|x|\geq 1$. Equivalently, $|1+x|^p\geq 1+px+a_p|x|^p$ for $|x|> 1$ (note that here I can replace $|x|\geq 1$ by $|x|>1$).
Consider the function $f:[-1,0)\cup (0,1]\to \mathbb{R}$ defined by $f(x)=\dfrac{|1+x|^p-1-px}{x^2}$. Clearly $f(x)$ is continuous and positive on $[-1,0)\cup (0,1]$.
Note that $\lim \limits_{x\to 0} f(x)=1$. Hence $\exists \delta \in (0,1)$ such that $\forall x\in [-\delta,0)\cup (0,\delta]$ we have $\frac{1}{2}<f(x)<\frac{3}{2}$.
Also on $[-1,-\delta]$ function $f(x)$ attains it minimum by Weierstrass say $f(\beta_1)$>0 and also on $[\delta,1]$ the function $f(x)$ attains its minimum say $f(\beta_2)>0$.
Hence for $0<|x|\leq 1$ we have $f(x)\geq \min \{f(\beta_1),f(\beta_2),\frac{1}{2}\}>0$. In other words, $\exists b_p>0$ such that $|1+x|^p\geq 1+px+b_p|x|^2$ for $|x|\leq 1$ (note that this is trivially true for $x=0$).
Denote $c_p:=\min \{a_p,b_p\}>0$ the we automatically obtain the desired result.
