Prove the inequality $|1+x|^p \ge 1+px+c_p\varphi_p(x)$

Question

Show that $|1+x|^p \ge 1+px+c_p\varphi_p(x)$, where $c_p$ is a constant depending only on $p$. $$\varphi_p(x) := \begin{cases}|x|^2 & |x| \le 1\\ |x|^p\ & |x| >1 \end{cases}$$ if $1<p\le2$, and $\varphi_p(x)=|x|^p$ on $\mathbb{R}$ if $p>2$.

This question is alike the binomial theorem $$(1+x)^p>1+px+\frac{p(p-1)}{2}x^2$$, for $p>1$ etc. But $\varphi_p(x)$ varies to the values of $x$ and $p$. How do I prove the above?

Answer 1

I think we have that $p \geqslant 1$.
By Bernoulli's inequality, we have that $\forall x\geqslant-1$: $$(1+x)^p \geqslant 1+px$$ And if $$0\geqslant1+px$$ We have that $$|1+x|^p\geqslant 0 \geqslant 1+px$$ So we need to work on the case when $-1\geqslant x\geqslant -\frac{1}{p}$. But since $p \geqslant 1$, we have that $\frac{1}{p} \leqslant 1$, so $-\frac{1}{p}\geqslant-1$.
So $\forall x \in \mathbb{R}$ we have that $$|1+x|^p \geqslant 1+px$$ Your $\varphi_p(x)\geqslant 0 \forall x$, so the inequality $$|1+x|^p \geqslant 1+px +c_p\varphi_p(x)$$ Will remain true $\forall c_p \leqslant0$.