$|x|^p+|y|^p\geq |x+y|^p$ for $0<p\leq 1$ [closed]

Question

How to prove such inequality: $|x|^p+|y|^p\geq |x+y|^p$ for $0<p\leq 1$ and $x,y \in \mathbb{R}$?

Answer 1

For $p=1$, we can prove this using the fact that $x^p$ is convex.

If $x$ or $y$ is $0$, then the result is obvious. For $0<p<1$, $x^p$ is concave on $(0, \infty)$. So for $x,y > 0$, $$\frac{y}{x+y}0^p + \frac{x}{x+y}(x+y)^p \le x^p $$$$\frac{x}{x+y}0^p + \frac{y}{x+y}(x+y)^p \le y^p $$

So for $x,y>0$$$x^p + y^p \ge (x+y)^p$$

And hence, (assuming the result with $p=1$) $\forall x,y \in \mathbb R$, $$|x+y|^p \le(|x| + |y|)^p \le|x|^p + |y|^p$$