A sum of inversion preserving summands using CS inequality Suppose that $p_k>0$ for $1\leq k\leq n$ and $\sum_{k=1}^np_k=1$. Show that 
$$\sum_{k=1}^n\left(p_k+\frac1{p_k}\right)^2\geq n^3+2n+\frac1n$$

Here's my try using Cauchy-Schwarz inequality:
$$\sum_{k=1}^n\left(p_k+\frac1{p_k}\right)^2\cdot\sum_{k=1}^np_k^2\geq\left(\sum_{k=1}^n\left(p_k^2+1\right)\right)^2=n^2+2n\sum_{k=1}^np_k^2+\left(\sum_{k=1}^np_k^2\right)^2$$
However, 
$$\sqrt{\frac{\sum_{k=1}^np_k^2}n}\geq\frac{\sum_{k=1}^np_k}n=\frac1n$$
Or,
$$\sum_{k=1}^np_k^2\geq\frac1n$$
Hence we have
$$\sum_{k=1}^n\left(p_k+\frac1{p_k}\right)^2\cdot\sum_{k=1}^np_k^2\geq n^2+2+\frac1{n^2}$$

I'm stuck here now.
 A: Perhaps you are making it yourself a bit difficult with Cauchy-Schwarz.
Expanding the LHS, we obtain $$\sum_{k=1}^n\left(p_k^2+2+\frac1{p_k^2}\right).$$
We'll estimate this in two steps. First, $\sum p_k^2\geq n\left(\frac{\sum p_k}n\right)^2$ by QM-AM. That is, $\sum p_k^2\geq\frac1n$.
Furthermore, $\sqrt{\frac{\sum\frac1{p_k^2}}n}\geq\frac n{\sum\frac1{\frac1{p_k}}}=n$ by QM-HM, which implies $\sum\frac1{p_k^2}\geq n^3$.
All this together gives $LHS\geq n^3+\frac1n+\displaystyle\sum_{k=1}^n2=n^3+2n+\frac1n$.

The main reason why your attempt failed is, I think, that you multiplied the LHS with $\sum p_k^2$. Since we are only given a restriction on $\sum p_k$, there is no nice upper bound for $\sum p_k^2$. Even if you arrived at a nice expression of the form
$$\left(\sum_{k=1}^n(p_k^2+1)\right)^2\geq f(n)$$
you still would have to divide both sides by $\sum p_k^2$. But as we can't find an upper bound for $\sum p_k^2$, we can't find a lower bound for $\frac{f(n)}{\sum p_k^2}$, which means the work would have been useless...
A: Noticed that $RHS=\dfrac{\left(n^2+1\right)^2}{n}$, We only need to prove
$$
n\sum_{k=1}^n \left(p_k+\frac{1}{p_k}\right)^2 \ge \left( n^2+1 \right) ^2
$$
Since $n = \displaystyle\sum_{k=1}^n 1$,
$$
\sum_{k=1}^n {1} \cdot \sum_{k=1}^n {\left(p_k+\frac{1}{p_k}\right)^2} \ge \left[ \sum_{k=1}^n {\left(p_k+\frac{1}{p_k}\right)} \right]^2 = \left( 1+\sum_{k=1}^n {\frac{1}{p_k}} \right)^2
$$
and
$$
\sum_{k=1}^n {\frac{1}{p_k}} = \sum_{k=1}^n {p_k} \cdot \sum_{k=1}^n {\frac{1}{p_k}} \ge \left( \sum_{k=1}^n \left({p_k\cdot\frac{1}{p_k}}\right) \right)^2 = n^2
$$
A: Let f(x(1),x(2),..,x(n)) = x(1)^2 + 1/x(1)^2 +...+ x(n)^2 + 1/x(n)^2 and Use Lagrange Multiplier with g(x(1), x(2),...,x(n)) = x(1) + (2) +...+ x(n) - 1 = 0. Taking partial derivatives lead to: 
2x(i) - 2/x(i)^3 = 2x(j) - 2/x(j)^3 for all i and j this gives x(i) = x(j) for all i and j. So x(1) = x(2) = ...= x(n) = 1/n. Thus Min f = f(1/n,1/n,..,1/n) = n^3 + 1/n. This implies the answer.
