Inequality: $0<\alpha<1$, $(\sum_{i=1}^na_i)^{\alpha} \leq \sum_{i=1}^na_i^\alpha$

Question

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?

Answer 1

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 $(a_{1}+a_{2})^{\alpha}=a_{1}^{\alpha}\left(1+\dfrac{a_{2}}{a_{1}}\right)^{\alpha}\leq a_{1}^{\alpha}\left(1+\left(\dfrac{a_{2}}{a_{1}}\right)^{\alpha}\right)=(a_{1}^{\alpha}+a_{2}^{\alpha})$.

Answer 2

Let $\;\displaystyle b_i = \frac{a_i}{\sum_{i=1}^na_i} \in (0,1)\;$ then $\,\sum_{i=1}^n b_i = 1\,$ and $\,b_i \lt b_i^\alpha\,$ for $\,0 \lt \alpha \lt 1\,$, therefore:

$$1 = \sum_{i=1}^n {b_i} \lt \sum_{i=1}^n {b_i}^\alpha \;\;\iff\;\; 1 \lt \sum_{i=1}^n \left(\frac{a_i}{\sum_{i=1}^na_i}\right)^\alpha \;\;\iff\;\; \left(\sum_{i=1}^na_i\right)^{\alpha} \lt \sum_{i=1}^na_i^\alpha$$

Answer 3

Hint Instead of Jensen, use the more general Karamata’s inequality, recognising $x^\alpha$ is concave and $(\sum a_i, 0,0,...0)\succ (a_1, a_2,...a_n)$