<s>[INST] 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.
 [/INST] 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
$$
\sum_{k=1}^n {\frac{1}{p_k}} = \sum_{k=1}^n {p_k} \cdot \sum_{k=1}^n {\frac{1}{p_k}} \ge \left( \sum_{k=1}^n \left({p_k\cdot\frac{1}{p_k}}\right) \right)^2 = n^2
$$
</s>