# Is $f(t)=t^\alpha$ for $\alpha\in(0,1)$ a sub-additive function? [duplicate]

Is the function

$$f(t)= t^{\alpha},\quad \alpha\in (0,1)$$

My teacher said categorically that this is true. But I'm not so sure.

EDIT: $~~~~~0\leq t\leq 1$

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Try with $0\le t\le 1$. –  Vittorio Patriarca Jun 5 '12 at 20:49
By homogeneity, in the inequality $(x+y)^{\alpha}\leq x^{\alpha}+y^{\alpha}$ just deal with the case $y=1$. Let $f(t):=t^{\alpha}+1-(t+1)^{\alpha}$. The derivative has the sign of $t^{\alpha-1}-(t+1)^{\alpha-1}$ which is non-negative since $\alpha<1$. Hence $f(t)\geq f(0)=0$ and we are done.
The notation $a^\alpha$ is a bit confusing, you should consider using $x,y$ or $u,v$ instead methinks. –  Asaf Karagila Jun 5 '12 at 21:00