TITLE: Show $\sum_{k=1}^n (p_k + \frac{1}{p_k})^2\geq n^3 + 2n + \frac{1}{n}$ ; $p_k\geq 0 \forall k$ and $\sum_kp_k=1$
QUESTION [5 upvotes]: Attempt: 
$\sum_{k=1}^n (p_k + \frac{1}{p_k})^2 = 2n + \sum_k p_k^2 + \sum_k \frac{1}{p_k^2}$
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.

REPLY [2 votes]: 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.