Inequality: $0<\alpha<1$, $(\sum_{i=1}^na_i)^{\alpha} \leq \sum_{i=1}^na_i^\alpha$ I would like to know if the following if true for $0<\alpha<1$, $a_i>0$ for all $i$.$$(\sum_{i=1}^na_i)^{\alpha} \leq \sum_{i=1}^na_i^\alpha$$
This looks like Jensen's inequality with counting measure. But $x^\alpha$ is 
a concave function, which means the direction should be reversed . However, I checked $\sqrt{(1+1)}\leq \sqrt1+\sqrt1$ which admits the same inequality. Can anybody prove or disprove the inequality?

For $n=2$, and $a>0$
\begin{align*}
(1+a)^{\alpha}-a^{\alpha}&=\int_{0}^{1}\dfrac{d}{dt}((1-t)a+t(a+1))^{\alpha}dt\\
&=\int_{0}^{1}\alpha((1-t)a+t(a+1))^{\alpha-1}dt\\
&=\int_{0}^{1}\alpha(a+t)^{\alpha-1}dt\\
&\leq\int_{0}^{1}\alpha t^{\alpha-1}dt\\
&=1,
\end{align*}
so $(1+a)^{\alpha}\leq 1+a^{\alpha}$.
So $(a_{1}+a_{2})^{\alpha}=a_{1}^{\alpha}\left(1+\dfrac{a_{2}}{a_{1}}\right)^{\alpha}\leq a_{1}^{\alpha}\left(1+\left(\dfrac{a_{2}}{a_{1}}\right)^{\alpha}\right)=(a_{1}^{\alpha}+a_{2}^{\alpha})$.
