is it true that, for $t \geq 0$: $\left|\frac{x+y}{2}\right|^t+\left|\frac{x-y}{2}\right|^t \le \frac{1}{2}|x|^t+\frac{1}{2}|y|^t $

Question

Is it the following inequality true for $t \geq 2$? And for $0\leq t \lt 2$?

$$\left|\frac{x+y}{2}\right|^t+\left|\frac{x-y}{2}\right|^t \le \frac{1}{2}|x|^t+\frac{1}{2}|y|^t $$

For $t = 2$ it is true because it's the parallelogram identity. For $t \gt 2$:

We have that $$|x+y|^t = 2^t\left|\frac{x+y}{2}\right|^t\le 2^t \left(\frac{1}{2}|x|^t+\frac{1}{2}|y|^t\right)$$ because of the convexity of $|*|^t$, hence: $$\left|\frac{x+y}{2}\right|^t\le \frac{1}{2}|x|^t+\frac{1}{2}|y|^t$$

So I would say that the answer is no, but maybe I am missing something.

Answer 1

$t=1,x=2,y=1$ is a counter-example (with $t<2$).

The inequality is true for $t \geq 2$: Let $|u|^{t}+|v|^{t} \leq 1$. Then ($|u|\leq 1$ and $|v|\leq 1$ so) $|\frac {u+v} 2|^{t}+|\frac {u-v} 2|^{t} \leq |\frac {u+v} 2|^{2}+|\frac {u-v} 2|^{2} =\frac {u^{2}+v^{2}} 2\leq 1$ . Now take $u=\frac x r, y=\frac y r$ where $r=\frac 1 {(|x|^{t}+|y|^{t})^{1/t}}$