# $|x|^p+|y|^p\geq |x+y|^p$ for $0<p\leq 1$

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

#### Solutions Collecting From Web of "$|x|^p+|y|^p\geq |x+y|^p$ for $0<p\leq 1$"

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\$, $$|x+y|^p \le(|x| + |y|)^p \le|x|^p + |y|^p$$