Inequality about $x^p$ for $p\geq 2$ a real number I am stuck with the following inequality
$$g^2(n+2)+2g(n+2)g(n)-g^2(n+1)-2g(n+1)g(n+2)\geq 0,$$
for all integer $n\geq 1$. Here, $g(x)=x^p$ where $p\geq 2$ is a real number. I need help.

It suffices to prove that
$$(n + 2)^{2p} + 2(n + 2)^p n^p - (n + 1)^{2p} - 2(n + 1)^p(n + 2)^p \ge 0$$
or
$$[(n + 2)^p - (n + 1)^p]^2
- 2(n + 1)^{2p} + 2(n + 2)^p n^p \ge 0$$
or (multiplying both sides by $(n + 1)^{-2p}$)
$$\left(\left(1 + \frac{1}{n +  1}\right)^p - 1\right)^2
- 2 + 2\left(1 - \frac{1}{(n + 1)^2}\right)^p \ge 0.$$
Using Bernoulli inequality, we have
$$\left(1 + \frac{1}{n +  1}\right)^p \ge 1 + \frac{p}{n + 1}$$
and
$$\left(1 - \frac{1}{(n + 1)^2}\right)^p \ge 1 - \frac{p}{(n + 1)^2}.$$
It suffices to prove that
$$\left(\frac{p}{n + 1}\right)^2
- 2 + 2\left(1 - \frac{p}{(n + 1)^2}\right) \ge 0$$
or
$$\frac{p(p - 2)}{(n + 1)^2} \ge 0$$
which is true.
We are done.
