Strict subadditivity of $x\mapsto x^t$ for $00$ Let $f:\mathbb (0,\infty)\to\mathbb R, x\mapsto x^t$ for some $0<t<1$. I want to know if $f$ is strictly subadditiv, i.e. $f(x)+f(y)>f(x+y)$.
If $t=\frac1k$ for some $k>0$, it's easy to see that $f$ is strictly subadditiv. I've checked the subadditivity of $f$ for certain values $t\neq\frac1k$ using the computer, so I am thinking that $f$ should be strictly subadditiv for all $t$. My question is how can you prove it for all $0<t<1$? I've tried to use the generalized binomial theorem but it didn't get me any futher. 

Yes, $f$ is strictly sub additive.
To prove it, fix some $x,y>0$ and observe that for $0<t<1$, we have $s^{t-1}>(x+s)^{t-1}$ for all $s>0$.
Integrating both side from $0$ to $y$, we get $\int_{0}^{y}s^{t-1}ds>\int_{0}^{y}(x+s)^{t-1}ds$, which gives $y^t>(x+y)^t-x^t$.
Hence the required sub additivity follows.
