<s>[INST] generalized f-mean/power mean inequalities I'd like to prove the following inequality (which seems to be true by numerics)
$$
(p-1)\frac{a^2+b^2}{2} \leq \Big(\frac{a^p+b^p}{2}\Big)^{2/p}
$$
for all $a,b\in [0,1]$ and $p\in [1,2]$. I'd eventually like a generalization when I am summing over many numbers and not just $2$, but I'd like to see how to prove this simple case first.
 [/INST] By applying the Holder's inequality  where $\frac{1}{m}+\frac{1}{n} = 1$
$$|x_1 y_1 +x_2y_2| \le (x_1^m +x_2^m)^{\frac{1}{m}} (y_1^m +y_2^n)^{\frac{1}{n}}$$
with $(m,n)=\left(\frac{p}{2},\frac{p}{p-2} \right)$, $(x_1,x_2) = (a^2,b^2)$  and $(y_1,y_2) = (1,1)$, we have
\begin{align}
a^2 +b^2 \le \left((a^2)^{\frac{p}{2}}+(b^2)^{\frac{p}{2}} \right)^{\frac{2}{p}} \left(1+1 \right)^{\frac{p-2}{p}} &\iff a^2 +b^2 \le \left(a^p +b^p\right)^{\frac{2}{p}} 2^{\frac{p-2}{p}} \\
&\iff \frac{a^2 +b^2}{2} \le \left(\frac{a^p +b^p}{2}\right)^{\frac{2}{p}} \tag{*}
\end{align}
And because $\frac{a^2 +b^2}{2} \ge (p-1)\frac{a^2 +b^2}{2} $ for $p \in [1,2]$, then from $(*)$ we can conclude that
$$
(p-1)\frac{a^2+b^2}{2} \leq \Big(\frac{a^p+b^p}{2}\Big)^{2/p}
$$
</s>