1
$\begingroup$

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.


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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks for the help, but I already have done the first part of showing the inequality. Rewrote the description to clarify that am currently having trouble showing equality implies each $p_k = 1/n$. $\endgroup$ – user246678 Feb 22 at 6:00
  • 1
    $\begingroup$ @user246678 I added something. See now. $\endgroup$ – Michael Rozenberg Feb 22 at 6:04
  • $\begingroup$ Thanks again, but I am unfamiliar with the $||$ symbol (in the context of vectors). What does it mean here? $\endgroup$ – user246678 Feb 22 at 6:16
  • 1
    $\begingroup$ It means that there is $\lambda$, for which $(p_1,p_2,...,p_n)=\lambda(1,1,...,1)$ or $p_k=\lambda,$ which gives $p_1=p_2=...=p_n$. $\endgroup$ – Michael Rozenberg Feb 22 at 6:19
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks for the help! If I wanted to show a proper proof that the equality implies each $p_1 = \cdots = p_n$ (i.e. without assuming the 3 inequalities were equalities, since it's true for $\leq$ case not necessarily for $\geq$), would it be possible to show this fairly easily? I was thinking of using proof by contradiction but wasn't sure how to approach it afterwards. $\endgroup$ – user246678 Feb 22 at 3:21
1
$\begingroup$

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 $p_1=p_2=\dots=p_n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.