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$\sum_{k=1}^n (p_k + \frac{1}{p_k})^2 = 2n + \sum_k p_k^2 + \sum_k \frac{1}{p_k^2}$

I used the Cauchy inequality to decompose 1 as $\sqrt{p_k}(\frac{1}{\sqrt{p_k}})$ and got

$n^2\leq \sum_k \frac{1}{p_k}$

I could, use the Cauchy inequality again on $\frac{1}{p_k}\cdot 1$ to get $(\sum\frac{1}{p_k})^2 < n \sum\frac{1}{p_k^2}$

$\Rightarrow n^4< (\sum\frac{1}{p_k})^2 < n \sum\frac{1}{p_k^2} $

$\Rightarrow n^3\leq\sum\frac{1}{p_k^2} $

what about $\sum p_k^2$. There's a mental block here. Any help would be appreciated.

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+1 for showing your working... – Aryabhata Mar 5 '11 at 23:27
up vote 3 down vote accepted

Let $$np' = \sum p_i$$

Consider $$\sum (p_i - p')^2$$

btw, I presume you meant $(\sum \frac{1}{p_k})^2$ instead of $\sum (\frac{1}{p_k})^2$

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i don't get it. – Please Delete Account Mar 5 '11 at 23:21
@Approximist: Expand it out and see what you get! – Aryabhata Mar 5 '11 at 23:23
yes changed it. – Please Delete Account Mar 5 '11 at 23:23
i see.. i can't beleive ive used that trick so many times in other contexts. thanks. oh and saw your profile.. badhiyan hai. – Please Delete Account Mar 5 '11 at 23:30
@Approx: Happens and shukriya :-) – Aryabhata Mar 5 '11 at 23:37

For reference, this problem is exercise 1.6 (page 14) of The Cauchy-Schwarz Master Class by J. Michael Steele. His solution is similar to Zarrax's, but for the lower bound on $\sum_{k=1}^n {1\over p_k}$ he uses the Cauchy-Schwarz inequality twice; first on the splitting $1=\sqrt{p_k}\cdot {1\over \sqrt{p_k}}$ and then on the splitting ${1\over p_k}=1\cdot {1\over p_k}$. Edit: I forgot to mention that the OP did this in the same way.

In Exercise 13.6 Steele gives a generalization, that for $\alpha>0$, we have $$\sum_{k=1}^n\left(p_k+{1\over p_k}\right)^\alpha \geq {(n^2+1)^\alpha\over n^{\alpha-1}}.$$

If you like inequalities, I strongly suggest getting this book. It is a gem!

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You have reduced the problem to showing that ${\displaystyle \sum_{k=1}^n p_k^2 + \sum_{k=1}^n {1 \over p_k^2} \geq {1 \over n} + n^3}$. Note that equality is achieved here when each $p_k = {1 \over n}$, when the first sum adds to ${1 \over n}$ and the second sum adds to $n^3$. Thus it makes sense to try to show that the first sum is at least ${1 \over n}$ and the second sum is at least $n^3$.

For the first sum, one can use the Cauchy-Schwarz inequality to conclude that $n \sum_{k=1}^n {p_k}^2 \geq (\sum_{k=1}^n {p_k})^2 = 1$, so that the first sum is in fact at least ${1 \over n}$. For the second sum, one similarly has $n \sum_{k=1}^n {1 \over p_k^2} \geq (\sum_{k=1}^n {1 \over p_k})^2$ or equivalently, $\sum_{k=1}^n {1 \over p_k^2} \geq {1 \over n} (\sum_{k=1}^n {1 \over p_k})^2$. To analyze $\sum_{k=1}^n {1 \over p_k}$, one may use the arithmetic-harmonic inequality, which says that for positive real numbers $a_1,...,a_n$ one has $${a_1 + ... + a_n \over n} \geq {n \over {1 \over a_1} + ... + {1 \over a_n}}$$ Applying this to $a_k = {1 \over p_k}$ gives that $$\sum_{k=1}^n {1 \over p_k} \geq n^2{1 \over \sum_{k=1}^n p_k}$$ $$ = n^2$$ So the second sum is at least ${1 \over n}(n^2)^2 = n^3$ as needed.

Edit: Naturally, in the time it took to write this out, someone already solved it.. such is life :)

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$f(p) = (p+1/p)$ is convex for $p > 0$. The sum is therefore maximized when the $p_i$ are equal.

The same argument handles $f(p)^a$ and $\log f(p)$ for $0 < p \leq 2$.

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I would say $f(x) = (x+1/x)^2$ is convex for $x \neq 0$. So $\sum \frac 1n f(p_i) \geq f(\sum \frac1n p_i)$. Now we find easily: $$ \sum_{i=1}^n \left(p_i+\frac1{p_i}\right)^2 \geq n \times f(1/n) = n \times \left(\frac1n+n\right)^2 $$ which is exactly what you want.

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+1 for nice try. – puresky Nov 16 '11 at 5:03

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