Another Fun Inequality

MarkFL

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For all real $0<a_i$, where $1\le i\le n$, prove:

$\displaystyle \sum_{k=1}^n\left(a_k^{-2}\right)\ge\sum_{k=1}^n\left(a_k^{2}\right)$

given that:

$\displaystyle \sum_{k=1}^n\left(a_k\right)=n$
 

tango

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For all real $0<a_i$, where $1\le i\le n$, prove:

$\displaystyle \sum_{k=1}^n\left(a_k^{-2}\right)\ge\sum_{k=1}^n\left(a_k^{2}\right)$

given that:

$\displaystyle \sum_{k=1}^n\left(a_k\right)=n$

That one is easy.

And God said:

Let
$\displaystyle \sum_{k=1}^n\left(a_k^{-2}\right)\ge\sum_{k=1}^n\left(a_k^{2}\right)$

given that:

$\displaystyle \sum_{k=1}^n\left(a_k\right)=n$

and it was so, and God saw that it was good.

What's next?
 
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