Q:

Let $x,y \in l_p$, proof that $2^k (||x||^p + ||y||^p)^{2/p} \leq 2 (||x||^2 + ||y||^2)$, where $k = 2 - 2/p$ and $1<p\leq 2$

Let $x,y \in l_p$, proof that $2^k (||x||^p + ||y||^p)^{2/p} \leq 2 (||x||^2 + ||y||^2)$, where $k = 2 - 2/p$ and $1<p\leq 2$ 
My attempt:
It's equivalent to proof the following inequality:
$$ \left ( \frac{||x||^p + ||y||^p}{2} \right )^{2/p} \leq \left ( \frac{||x||^2 + ||y||^2}{2} \right ) \iff \left ( \frac{||x||^p + ||y||^p}{2} \right )^{2} \leq \left ( \frac{||x||^2 + ||y||^2}{2} \right )^p $$
As $1 < p \leq 2 \to 2^p \leq 2^2 \to \frac{1}{2^2} \leq \frac{1}{2^p}$. So it's sufficient showing that $$(||x||^p + ||y||^p)^2 \leq (||x||^2 + ||y||^2)^p$$
I think I should use Holder's inequality, but I can't see where.
As $||x||^p = \sum_{n \in \mathbb N} |x_n|^p$ and $||y||^p = \sum_{n \in \mathbb N} |y_n|^p$.
$(||x||^p + ||y||^p)^2 = (\sum_{n \in \mathbb N} |x_n|^p + \sum_{n \in \mathbb N} |y_n|^p )^2 =  (\sum_{n \in \mathbb N} |x_n|^p )^2 + (\sum_{n \in \mathbb N} |y_n|^p )^2  + \\ + 2\sum_{n \in \mathbb N} |x_n|^p\sum_{n \in \mathbb N} |y_n|^p$
As
$$(\sum_{n \in \mathbb N} |x_k|^p )^2 = (|x_1|^p + |x_2|^p + |x_3|^p + ...)^2 = \sum_{n \geq 1} |x_n|^{2p} + 2\sum_{m \geq 1} \left ( \sum_{n \geq 1} |x_n x_{n+m}|^p \right ) $$
$$ (\sum_{n \in \mathbb N} |y_k|^p )^2 =   \sum_{n \geq 1} |y_n|^{2p} + 2\sum_{m \geq 1} \left ( \sum_{n \geq 1} |y_n y_{n+m}|^p \right ) $$
$$2 \sum_{n \in \mathbb N} |x_n|^p\sum_{n \in \mathbb N} |y_n|^p  = 2\sum_{n \geq 1} |x_n y_n|^p + 2 \sum_{m \geq 1} \left ( \sum_{n \geq 1} |x_n y_{n+m}|^p + \sum_{n \geq 1} |x_{n+m} y_n|^p \right )$$
Suming the three equations
$$ (||x||^p + ||y||^p)^2 = \left ( \sum_{n \geq 1} (|x_n|^{2p} + |y_n|^{2p} + 2 |x_n y_n|^p) \right ) + \\ 2\sum_{m \geq 1} \left ( \sum_{n \geq 1} |x_n x_{n+m}|^p  + \sum_{n \geq 1} |y_n y_{n+m}|^p  + \sum_{n \geq 1} |x_n y_{n+m}|^p + \sum_{n \geq 1} |x_{n+m} y_n|^p \right )  $$
Well, I don't know how to apply Holder's inequality in a way that it works, besides there is another way in doing this question?
Thank you

A:

denote $X=||x||$ and $Y=||y||$, then your inequality becomes $$2^{2-\frac{2}{p}}(X^p+Y^p)^{\frac{2}{p}}\le 2(X^2+Y^2)$$ or equivalently $$\sqrt[p]{\frac{X^p+Y^p}{2}}\le\sqrt[2]{\frac{X^2+Y^2}{2}}$$ what is just an inequality between power means
https://www.artofproblemsolving.com/wiki/index.php?title=Power_Mean_Inequality

