Given that $p_k> 0$ and $p_1+p_2+\cdots+p_n=1$, prove that \begin{equation} \sum_{k=1}^n \left(p_k+\frac{1}{p_k}\right)^2\geq n^3+2n+\frac{1}{n}. \end{equation}
I believe that Cauchy's inequality should be used at some point but I haven't figured out how. Expanding the square on the left-hand side gives $2n$ immediately, but I have problem producing the cubic term and $\frac{1}{n}$. Could anyone please offer some insight? It is ok to use any method, not necessarily Cauchy's inequality.
Let $f(x) = \left(x+\dfrac{1}{x}\right)^2 = x^2+2+\dfrac{1}{x^2}$. Then, $f''(x) = 2+\dfrac{6}{x^4} > 0$ for all $x \in [0,1]$.
Since $f$ is convex on $[0,1]$, for any $p_1,\ldots,p_n \in [0,1]$ where $p_1+\cdots+p_n = 1$, we can use Jensen's Inequality to get $$\dfrac{1}{n}\sum_{k = 1}^{n}f(p_k) \ge f\left(\dfrac{1}{n}\sum_{k = 1}^{n}p_k\right)$$ $$\dfrac{1}{n}\sum_{k = 1}^{n}\left(p_k+\dfrac{1}{p_k}\right)^2 \ge f\left(\dfrac{1}{n}\right)$$ $$\sum_{k = 1}^{n}\left(p_k+\dfrac{1}{p_k}\right)^2 \ge nf\left(\dfrac{1}{n}\right) = n^3+2n+\dfrac{1}{n}.$$
Note that $$\sum_{k=1}^n \left(p_k+\frac{1}{p_k}\right)^2 =\sum_{k=1}^n p_k^2+2n+\sum_{k=1}^n \frac{1}{p_k^2}$$
Now note that $$\left(\sum_{k=1}^n p_k^2\right)\left(\sum_{k=1}^n 1\right) \ge \left(\sum_{k=1}^n p_k\right)^2 \implies \sum_{k=1}^n p_k^2 \ge \frac{1}{n} $$$$\left(\sum_{k=1}^n \frac{1}{p_k^2}\right)\left(\sum_{k=1}^n p_k\right)\left(\sum_{k=1}^n p_k\right) \ge \left(\sum_{k=1}^n 1\right)^3 \implies \sum_{k=1}^n \frac{1}{p_k^2} \ge n^3$$
From Hölder's inequality.
Adding these two inequalities, we are done.
$\sum_{k=1}^n \left(p_k+\frac{1}{p_k}\right)^2 = \sum_{k=1}^n \left(p_k^2 + \frac{1}{p_k^2}+2\right)=\sum_{k=1}^n p_k^2 + \sum_{k=1}^n \frac{1}{p_k^2}+2n\,$. By AM-GM:
$\;\;\sqrt{\cfrac{1}{n}\sum_{k=1}^n p_k^2} \;\ge\; \cfrac{1}{n} \sum_{k=1}^n p_k = \cfrac{1}{n} \quad\implies\quad \sum_{k=1}^n p_k^2 \;\ge\; \cfrac{1}{n}$
$\;\;\sqrt{\cfrac{n}{\sum_{k=1}^n \cfrac{1}{p_k^2}}} \;\le\; \cfrac{1}{n} \sum_{k=1}^n p_k = \cfrac{1}{n} \quad\implies\quad \sum_{k=1}^n \cfrac{1}{p_k^2} \ge n^3$
Adding up the above gives the stated inequality and, since it's all based on AM-GM, the equality holds iff all $p_k$ are equal.
The question is from chapter 1 of J. Michael Steele's book ``The Cauchy-Schwarz Master Class''. I didn't know that the book has solutions at the back. Below is the author's solution, which only employs Cauchy's inequality.
First, we expand the square. \begin{equation} \sum_{k=1}^n \left(p_k+\frac{1}{p_k}\right)^2 = \sum_{k=1}^n p_k^2 + 2n + \sum_{k=1}^n \frac{1}{p_k^2}. \end{equation}
Consider the first sum. Notice that \begin{equation} \left(\sum_{k=1}^n p_k^2 \right) \left(\sum_{k=1}^n 1\right) \geq \left(\sum_{k=1}^n p_k\right)^2=1, \end{equation} from which we obtain \begin{equation} \sum_{k=1}^n p_k^2 \geq \frac{1}{n}. \end{equation}
Now, consider the second sum, for which we first consider the sum $\sum_{k=1}^n\frac{1}{p_k}$. Observe that \begin{equation} \sum_{k=1}^n 1 = \sum_{k=1}^n \sqrt{p_k}\frac{1}{\sqrt{p_k}} \leq \left(\sum_{k=1}^n p_k\right)^{\frac{1}{2}} \left(\sum_{k=1}^n \frac{1}{p_k} \right)^{\frac{1}{2}}. \end{equation} It follows that \begin{equation} \sum_{k=1}^n \frac{1}{p_k} \geq n^2. \end{equation}
Next, we can obtain a lower bound for $\sum_{k=1}^n\frac{1}{p_k^2}$. By Cauchy's inequality, \begin{equation} \sum_{k=1}^n\frac{1}{p_k} = \sum_{k=1}^n\frac{1}{p_k} 1 \leq \left(\sum_{k=1}^n\frac{1}{p_k^2}\right)^{\frac{1}{2}} \left(\sum_{k=1}^n 1^2\right)^{\frac{1}{2}}. \end{equation} Hence, \begin{equation} \left(\sum_{k=1}^n\frac{1}{p_k^2}\right)^{\frac{1}{2}} \left(\sum_{k=1}^n 1^2\right)^{\frac{1}{2}} \geq \sum_{k=1}^n\frac{1}{p_k} \geq n^2. \end{equation} Squaring both sides, we obtain \begin{equation} \left(\sum_{k=1}^n\frac{1}{p_k^2}\right) n \geq n^4. \end{equation} Therefore, \begin{equation} \sum_{k=1}^n\frac{1}{p_k^2} \geq n^3. \end{equation}
The conclusion that \begin{equation} \sum_{k=1}^n \left(p_k+\frac{1}{p_k}\right)^2 \geq n^3 + 2n + \frac{1}{n} \end{equation} follows.