# Prove the inequality.

$$\left(\sum_{i=1}^na_i^p\right)^{1/p} \ge \left(\sum_{i=1}^na_i^q\right)^{1/q}$$ if $0 < p \le q$ for $a_i\ge 0$. I have proved that the inequality holds for $p=q$ (trivial) and i have also proved that it holds if one of the two sums is equal to 1, but I don't know how to continue. Please help.

Hint: use the Jensen inequality with $x\to x^{q/p}$.
• This says that, is $f$ is convex ($f''\ge 0$) then for every $c_i>0, d_i$ if $\sum c_i = 1$ then $f(\sum c_i d_i) \le \sum c_i f(d_i)$. – mookid Nov 2 '14 at 15:21
Set $b_i=a_i^q$ and $t=p/q\leq1$, then enough to show $$\sum_{i=1}^nb_i^t\geq (\sum_{i=1}^nb_i)^t.$$ i.e. $$\sum_{i=1}^n\left(\frac{b_i}{\sum_{i=1}^nb_i}\right)^t\geq1.$$ Note that $$\sum_{i=1}^n\left(\frac{b_i}{\sum_{i=1}^nb_i}\right)^t\geq\sum_{i=1}^n\left(\frac{b_i}{\sum_{i=1}^nb_i}\right)=1.$$ The first inequality is becuase $x^t\geq x$ for $t\leq 1$ and $x\leq 1$.