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.
Best Answer
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 $$\dfrac{1}{n}\sum_{k = 1}^{n}f(p_k) \ge f\left(\dfrac{1}{n}\sum_{k = 1}^{n}p_k\right)$$ $$\dfrac{1}{n}\sum_{k = 1}^{n}\left(p_k+\dfrac{1}{p_k}\right)^2 \ge f\left(\dfrac{1}{n}\right)$$ $$\sum_{k = 1}^{n}\left(p_k+\dfrac{1}{p_k}\right)^2 \ge nf\left(\dfrac{1}{n}\right) = n^3+2n+\dfrac{1}{n}.$$