Showing that if $p_1 + \cdots p_n = 1$ then $\displaystyle \sum_{k=1}^n \left(p_k + \dfrac {1}{p_k} \right)^2 \ge n^3+2n+\dfrac 1n$? This problem is from the book "Cauchy-Schwarz Masterclass":

Show that if $p_1 + \cdots p_n = 1$ with each $p_i$ positive, then $\displaystyle \sum_{k=1}^n \left(p_k + \dfrac {1}{p_k} \right)^2 \ge n^3+2n+\dfrac 1n$

I expanded the LHS and arrived at 
$$(p_1^2 + \cdots + p_n^2) + \left(\dfrac {1}{p_1^2} + \cdots +  \dfrac {1}{p_n^2}\right) \ge n^3 + \dfrac 1n$$
I was able to show that $p_1^2 + \cdots + p_n^2 \ge \dfrac 1n$ by applying C-S to the sequences $(p_1, ..., p_n)$ and $(p_1, ..., p_n)$. I also think that $\dfrac {1}{p_1^2} + \cdots +  \dfrac {1}{p_n^2} \ge n^3$ is true (because equality holds for $p_i = \dfrac 1n$ and I checked numerically for some values) but I am not able to prove it.
At this point in the book we only proed that classic $C-S$ inequality and the $C-S$ for inner product space, so I am hoping there is a solution only with these tools.
 A: Assume all $p_i$ are positive, otherwise the LHS is infinite and the inequality is trivial. Consider the function $f\colon(0,1)\to\mathbb{R}$ defined by
$$
f(x) = \left(x+\frac{1}{x}\right)^2,\qquad x\in(0,1) \tag{1}
$$
which is easily seen to be convex. (For instance, $f''(x) = 2+6/x^4 > 0$.)
By Jensen's inequality,
$$
\sum_{i=1}^n \frac{1}{n} f(p_i) \geq f\left(\sum_{i=1}^n \frac{p_i}{n}\right) = f\left(\frac{1}{n}\right) \tag{2}
$$
i.e.,
$$
\frac{1}{n} \sum_{i=1}^n \left(p_i+\frac{1}{p_i}\right)^2 \geq \frac{1}{n^2}+2 + n^2\,. \tag{3}
$$
Multiply both sides by $n$ to obtain the desired inequality.
A: In fact we seek to minimize $$\dfrac{1}{p_1^2}+\dfrac{1}{p_2^2}+\cdots+\dfrac{1}{p_n^2}$$respect to $$p_1+p_2+\cdots+p_n=1$$using Lagrange multipliers we get$$-\dfrac{2}{p_1^3}=\lambda\\-\dfrac{2}{p_2^3}=\lambda\\.\\.\\.\\-\dfrac{2}{p_n^3}=\lambda$$which yields to $$p_1=p_2=\cdots=p_n=\dfrac{1}{n}$$ and leads to the same result by substitution
A: This is a proof that only uses Cauchy Schwarz, the second part can be simplified if you are allowed to use AM $\ge$ HM.
Apply CS to $n$ copies of $1$ and $\displaystyle\;p_k+\frac{1}{p_k}$, we obtain
$$n \sum_{k=1}^n \left(p_k + \frac{1}{p_k}\right)^2
 = \left( \sum_{k=1}^n 1^2\right)\sum_{k=1}^n\left(p_k + \frac{1}{p_k}\right)^2
\ge \left(\sum_{k=1}^n p_k + \frac{1}{p_k}\right)^2
 = \left(1 + \sum_{k=1}^n \frac{1}{p_k}\right)^2$$
Apply CS again to $\sqrt{p_k}$ and $\displaystyle\;\frac{1}{\sqrt{p_k}}$, we obtain
$$\sum_{k=1}^n \frac{1}{p_k} = \sum_{k=1}^n \sqrt{p_k}^2 \sum_{k=1}^n \frac{1}{\sqrt{p_k}^2}
\ge \left(\sum_{k=1}^n \frac{\sqrt{p_k}}{\sqrt{p_k}}\right)^2 = n^2$$
Combine these two inequalities, we obtain
$$\sum_{k=1}^n \left(p_k + \frac{1}{p_k}\right)^2
\ge \frac1n \left(1 + n^2\right)^2 = n^3 + 2n + \frac1n$$
A: The Lagrangian reads
$$
L(p,\lambda) =\sum_{k=1}^n\left(p_k+\frac{1}{p_k}\right)^2+\lambda\left(\sum_{k=1}^n p_k - 1\right)
$$
The stationary onditions are
$$
2p_k-\frac{2}{p_k^3}+\lambda = 0, \;\; \forall k\\
\sum_{k=1}^n p_k = 1
$$
so we conclude $p_1=p_2=\cdots= p_n = \frac 1n$ hence
$$
\sum_{k=1}^n\left(p_k+\frac{1}{p_k}\right)^2 = n\left(n+\frac 1n\right)^2 = n^3+2n+\frac 1n
$$
