Is the operator $(\cdot)^{\frac{1}{p}}$ subadditive for $1<p<\infty$?
I am aware that the square root operator is subadditive for positive numbers, as for $a,b > 0$:
$\sqrt{a+b}\leq \sqrt{a}+\sqrt{b}\iff a+b\leq a+2\sqrt{a}\sqrt{b}+b$
and the last statement is trivially true since $\sqrt{a}\sqrt{b}\geq 0$.
How can I generalize this proof for $p\in (1,\infty)$?
Is the assumption on non-negativity of $a$ and $b$ necessary?
