Generalization of an inequaliy related to Cauchy-Schwarz

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.

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

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