<s>[INST] Proof of an inequality using Cauchy's inequality: $\sum_{k=1}^n \left(p_k+\frac{1}{p_k}\right)^2\geq n^3+2n+\frac{1}{n}$ 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.
 [/INST] 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.
</s>