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$$\left(\sum_{i=1}^na_i^p\right)^{1/p} \ge \left(\sum_{i=1}^na_i^q\right)^{1/q} $$ 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.

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Hint: use the Jensen inequality with $x\to x^{q/p}$.

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  • $\begingroup$ i have never heard of the jensens inequality $\endgroup$ – Baquesh Nov 2 '14 at 15:20
  • $\begingroup$ Do you know what is a convex function ? $\endgroup$ – Pierre Alvarez Nov 2 '14 at 15:21
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    $\begingroup$ This says that, is $f$ is convex ($f''\ge 0$) then for every $c_i>0, d_i$ if $\sum c_i = 1$ then $f(\sum c_i d_i) \le \sum c_i f(d_i)$. $\endgroup$ – mookid Nov 2 '14 at 15:21
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    $\begingroup$ ok thanks, ill try $\endgroup$ – Baquesh Nov 2 '14 at 15:24
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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 $x\leq 1$.

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  • $\begingroup$ but where do you get from here $\endgroup$ – Baquesh Nov 2 '14 at 18:44
  • $\begingroup$ @Baquesh what do you mean? I proved the inequality you asked. $\endgroup$ – Chen Jiang Nov 3 '14 at 6:01

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