# Is $x^t$ subadditive for $t \in [0,1]$?

I feel like I have seen this many times, that $(x+y)^t \leq x^t + y^t$ when $t \in [0,1]$, but I don't remember ever proving it. I feel like it surely must be true. Can someone lead me in the right direction on this one?

We assume that $x$ and $y$ are $\ge 0$, else exponentiation poses difficulties. And we might as well suppose that $y$ is positive. We also assume that $0\lt t\lt 1$, since the cases $t=0$ and $t=1$ are easy.
Fix $y$, and let $f(x)=(x+y)^t-x^t-y^t$. Then $$f'(x)=t\left(\frac{1}{(x+y)^{1-t}}-\frac{1}{x^{1-t}}\right).$$ The derivative is $\le 0$, so $f(x)$ is non increasing. Since $f(0)=0$, the result follows.
Or just note that $a^t\ge a$ for $0\le a,t\le 1$, denote $a=\frac x{x+y}$, $b=\frac y{x+y}$, and write $a^t+b^t\ge a+b=1$.