Is it true that for $p\in (1,2)$ the following inequalities holds: $ 2^{p-1} (|x|^p+|y|^p)\leq |x+y|^p+|x-y|^p \leq 2 (|x|^p+|y|^p)$ for $x, y \in \mathbb{R}$ ?
Thanks.
Is it true that for $p\in (1,2)$ the following inequalities holds: $ 2^{p-1} (|x|^p+|y|^p)\leq |x+y|^p+|x-y|^p \leq 2 (|x|^p+|y|^p)$ for $x, y \in \mathbb{R}$ ?
Thanks.
We first show that $x_1^p+x_2^p\geq (x_1^2+x_2^2)^{p/2}$ for non-negative $x_1,x_2$. By homogeneity it's enough to show that $f(t):=x^p+1-(x^2+1)^{p/2}$ is non-negative. Its derivative is $f'(x)=px^{p-1}-\frac p22x(x^2+1)^{p/2-1}\geq px^{p-1}-pxx^{p-2}=0$ since the map $s\mapsto s^{p/2-1}$ is decreasing ($p<2$). Now we apply it to $x_1=\frac{x+y}2$ and $x_2=\frac{x-y}2$. We get $\left|\frac{x+y}2\right|^p+\left|\frac{x-y}2\right|^p\geq\left(\left|\frac{x+y}2\right|^2+\left|\frac{x-y}2\right|^2\right)^{p/2}=\left(\frac{x^2}2+\frac{y^2}2\right)^{p/2}$ and since the map $t\mapsto t^{p/2}$ is concave, we get the first inequality.
To get the second one, put $u=x+y$ and $v=x-y$ and apply the first inequality to $u$ and $v$. Since $x=\frac{u+v}2$ and $y=\frac{u-v}2$ we have $2^{p-1}\left(\left|\frac{u+v}2\right|^p+\left|\frac{u-v}2\right|^p\right)\leq |u|^p+|v|^p$ so $2^{p-1}(|x|^p+|y|^p)\leq 2^p(|x+y|+|x-y|)$ and we are done.
For $p\geq 2$, the first inequality is reversed.