Finding equality of inequality via Cauchy-Schwarz

Question

Assuming $p_k > 0$, $1 \leq k \leq n$ and $p_1 + p_2 + \cdots + p_n = 1$, show that:

$$\sum_{k=1}^n \left( p_k + \frac{1}{p_k} \right)^2 \geq n^3 + 2n + 1/n$$

and determine necessary and sufficient conditions for equality to hold.

This comes from Michael Steele's 'Cauchy-Schwarz Master Class' book, exercise 1.6. I was able to solve the first part using Cauchy-Schwarz multiple times:

$$p_1 + \cdots + p_n = 1 \leq \sqrt{n} \left( \sum_{k=1}^n p_k ^2 \right)^{1/2}$$

$$\frac{1}{p_1} + \cdots + \frac{1}{p_n} \leq \sqrt{n} \left( \sum_{k=1}^n \frac{1}{p_k ^2} \right)^{1/2}$$

$$n \leq \left( \sum_{k=1}^n \frac{1}{p_k} \right)^{1/2}$$

using these 3 inequalities leads to the inequality.

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However for the second part, I'm not sure how Cauchy Schwarz (or some other technique, however this is the 1st chapter in the book) could be used to show instead:

$$\sum_{k=1}^n \left( p_k + \frac{1}{p_k} \right)^2 = n^3 + 2n + 1/n$$

I was aware (as well an answer here) that the above equality holds iff $p_1 = ... = p_n$ but am not sure how to prove that the equality implies that $p_1, p_2, \cdots = 1/n$. Any hints would be greatly helpful.

Answer 1

By C-S $$\sum_{k=1}^n\left(p_k+\frac{1}{p_k}\right)^2=\frac{1}{n}\sum_{k=1}^n1^2\sum_{k=1}^n\left(p_k+\frac{1}{p_k}\right)^2\geq\frac{1}{n}\left(\sum_{k=1}^n\left(p_k+\frac{1}{p_k}\right)\right)^2=$$ $$=\frac{1}{n}\left(1+\sum_{k=1}^np_k\sum_{k=1}^n\frac{1}{p_k}\right)^2\geq\frac{1}{n}(1+n^2)^2=n^3+2n+\frac{1}{n}.$$ The equality occurs, when $$(\sqrt{p_1},\sqrt{p_2},...,\sqrt{p_n})||\left(\frac{1}{\sqrt{p_1}},\frac{1}{\sqrt{p_2}},...,\frac{1}{\sqrt{p_n}}\right)$$ or $$(p_1,p_2,...,p_n)||(1,1,...1),$$

which with $p_1+p_2+...+p_n=1$ gives $p_1=p_2=...=p_n=\frac{1}{n}.$

By the way, our inequality is obviously true by Jensen.

Answer 2

Just write down the equality conditions for your three applications of Cauchy-Schwarz.

The first two inequalities are equalities if and only if $p_1 = \cdots = p_n$.

The third inequality is equality if and only if $p_1^2 = \cdots = p_n^2$, which is redundant given the above condition.

Answer 3

Use Jensen's inequality with the uniform measure on $\{1,\dots,n\}$.

The function $x \to (x+ 1/x)^2$ is convex. Therefore,

$$n \sum_{j=1}^n \frac{1}{n} (p_j + p_j^{-1})^2 \ge n (\sum_{j=1}^n \frac{p_j}{n} + (\sum_{j=1}^n \frac{p_j}{n})^{-1})^2= n ( \frac{1}{n} + n)^2=n ( \frac{1}{n^2} + 2n + n^2).$$ This gives the inequality. Equality in Jensen holds if and only if $p_1=p_2=\dots=p_n$.