Prove $(1+x)^p+(1-x)^p \ge 2(1+x^p)$ for $0\le x\le1$ and real number $p\ge2$. I don't know how to prove the following questions: If $p\ge2$ is real, then
$$
(1+x)^p+(1-x)^p \ge 2(1+x^p) \quad \text{for } 0\le x\le1;
$$
if $1\le p<2$, then opposite direction of the inequality holds.
I try to find a convex function but I could not. And I take the derivative of function $f(x):=(1+x)^p+(1-x)^p-2x^p$ but it still doesn't work.
EDIT
I tried the "Binomial series", but it could only be workable for the case $p\ge2$ but not for the case $1\le p<2$.


Lemma. for $f(x)=x^k$. if $k\geq 1$ then for $a,b>0$,then $a^k+b^k\leq (a+b)^k$. if $0\leq k\leq 1$, then $a^k+b^k\geq (a+b)^k$.

proof.when $k\geq 1$, then we have $(\frac{a}{a+b})^k\leq\frac{a}{a+b}$ and $(\frac{b}{a+b})^k\leq\frac{b}{a+b}$ for $\frac{a}{a+b}\leq 1$, $\frac{b}{a+b}\leq 1. $
Hence:$(\frac{a}{a+b})^k+(\frac{b}{a+b})^k\leq \frac{a}{a+b}+\frac{b}{a+b}=1$,which implies:$a^k+b^k\leq (a+b)^k$. In the similar way, we can prove if $0\geq k\leq 1$, then $a^k+b^k\geq (a+b)^k$.
Back to the problem:Set $f(x)= (1+x)^p+(1-x)^p-2(1+x^p)$,we have:
$$
f'(x)= p\times (1+x)^{p-1}-p\times (1-x^{p-1})-2p\times x^{p-1}
$$
When $p\geq 2$, then $p-1 \geq 1$.
Then for $0\leq x \leq 1$:
$$
(1+\frac{1}{x})^{p-1}\geq 2^{p-1}+(\frac{1}{x}-1)^{p-1}\geq 2+(\frac{1}{x}-1)^{p-1}
$$
which implies $f'(x)\geq 0$
Thus $f(x)\geq f(0)=0$
Applying the same method, when $1\leq p <2$, we have:
$$
f(x)\leq f(1)=2^p-2\times (1+1)\leq 0.
$$
