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For what values of $s\in [0,\infty),$ the function given by $f(x)=x^s$ sub-additive?

i.e. $f(a+b)\leq f(a)+f(b) \quad \forall a,b \in [0, \infty)$

I feel $f$ is sub-additive for $s\in [0,1]$ and not sub-additive for $s>1$

If so, How to prove it?

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For $s >1$ just take $a=b=1$ to see that it is not subadditive. For $s<1$ consider the function $f(a)=a^{s}+b^{s}-(a+b)^{s}$ with $b$ fixed. Note that $f'(a)=sa^{s-1}-s(a+b)^{s-1}$. Since $s-1 <0$ this shows that $f'(a)>0$ so $f$ is increasing. Also $f(0)=0$ so $f(a) \geq 0$ for all $a \geq 0$.

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