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A classic Cauchy-Schwarz problem is to show that if $p_1, p_2 \cdots p_k$ are positive real numbers with $p_1 + p_2 \cdots +p_k=1$ then $\sum_{k=1}^n (\frac{1}{p_k} +p_k)^2 \geq n^3 +2n + \frac{1}{n}$ with equality if and only if $p_i=\frac{1}{n}$ for all $i$.

It is also not hard to use Cauchy-Schwartz to show that if $p_1, p_2 \cdots p_k$ are positive real numbers with $p_1 + p_2 \cdots +p_k=1$ then $\sum_{k=1}^n (\frac{1}{p_k} +p_k) \geq n^2 +1$.

Taken together, these suggest the following generalization, which I'm hoping to see a proof or disproof of. If $p_1, p_2 \cdots p_k$ are positive real numbers with $p_1 + p_2 \cdots +p_k=1$

$\sum_{k=1}^n (\frac{1}{p_k} +p_k)^m \geq n(\frac{1}{n} +n)^m$ with equality only when the $p_i$ are all equal.

The general inequality is easy to see for $n=2$, from just looking at the function $f(x)=(\frac{1}{x}+x)^m$, since its derivative is zero at $x=1/2$ and then one can apply the first derivative test.

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  • $\begingroup$ Use Jensen's inequality $\endgroup$
    – kabenyuk
    Commented Dec 22, 2023 at 14:19
  • $\begingroup$ @kabenyuk That was my guess but it wasn't clear to me what function to use there. What function do you suggest? $\endgroup$
    – JoshuaZ
    Commented Dec 22, 2023 at 14:58
  • $\begingroup$ I think the function you wrote in the last lines of your question will do. $\endgroup$
    – kabenyuk
    Commented Dec 22, 2023 at 15:24
  • $\begingroup$ @kabenyuk Ah yes, you are correct! If you make that an answer I'll accept it. $\endgroup$
    – JoshuaZ
    Commented Dec 22, 2023 at 15:51
  • $\begingroup$ Happy to post my answer. $\endgroup$
    – kabenyuk
    Commented Dec 22, 2023 at 16:19

1 Answer 1

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Hint.

Let's use Jensen's inequality for the function $$ f(x)=\left(x+\frac{1}{x}\right)^m. $$

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