For $a, b \geq 0$, $0 < x < 1$, show $(a+b)^x \leq a^x + b^x$
Here is another way:
WLOG we may normalize with $a+b = 1$, so $a, b \in (0, 1) \implies a^x > a,\quad b^x > b$, so $a^x+b^x > a+ b = 1$.
For $a, b \geq 0$, $0 < x < 1$, show $(a+b)^x \leq a^x + b^x$
Here is another way:
WLOG we may normalize with $a+b = 1$, so $a, b \in (0, 1) \implies a^x > a,\quad b^x > b$, so $a^x+b^x > a+ b = 1$.