The normed space $\ell_p$ is UC

Question

I need to prove that $\ell_p$ ($p\in[2,+\infty)$) is Uniformly Convex (UC). The only thing I've left to do is to see that:

\begin{equation} ||x+y||_p^p+||x-y||_p^p\leq 2^{p-1}\cdot(||x||_p^p+||y||_p^p). \end{equation}

Using this is very easy to prove that $\delta_{\ell_p}(\varepsilon)=1-\left(1-\left(\dfrac{\varepsilon}{2}\right)^p\right)^{1/p}>0$ and deduced that $\ell_p$ is UC.

To prove the inequality above, is sufficient to demonstrate the inequality for real numbers rather than $p$-summable sequences.

What I've tried is the following.

For $x=0$, the inequality is trival. Suppose $x\neq 0$. Divinding both sides by $|x|^p$ we only need to prove that:

$$|1+z|^p+|1-z|^p\leq 2^{p-1}\cdot(1+|z|^p).$$

where z=y/x. If the inequality is true for $z$, then is true for $-z$. So, we can suppose that $z>0$. If $z=1$, again, the inequality is trivial. If $z>1$, then you can divide both sides by $z^p$ and obtain the same inequality for $w=1/z<1$, that is to say, we need to prove that:

$$(1+w)^p+(1-w)^p\leq 2^{p-1}\cdot(1+w^p),\forall w\in(0,1).$$

I called $f(w)=(1+w)^p+(1-w)^p-2^{p-1}\cdot(1+w^p)$ $w\in[0,1]$ and saw that $f(0)=2-2^{p-1}\leq 0$ and $f(1)=0$. Graphing the function, one can see that $f$ is increasing, so I have to prove that $f'\geq 0$ which is a bit difficult and I don't know how to do it.