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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. |
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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. |
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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 $$\dfrac{1}{n}\sum_{k = 1}^{n}f(p_k) \ge f\left(\dfrac{1}{n}\sum_{k = 1}^{n}p_k\right)$$ $$\dfrac{1}{n}\sum_{k = 1}^{n}\left(p_k+\dfrac{1}{p_k}\right)^2 \ge f\left(\dfrac{1}{n}\right)$$ $$\sum_{k = 1}^{n}\left(p_k+\dfrac{1}{p_k}\right)^2 \ge nf\left(\dfrac{1}{n}\right) = n^3+2n+\dfrac{1}{n}.$$ |
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$\sum_{k=1}^n \left(p_k+\frac{1}{p_k}\right)^2 = \sum_{k=1}^n \left(p_k^2 + \frac{1}{p_k^2}+2\right)=\sum_{k=1}^n p_k^2 + \sum_{k=1}^n \frac{1}{p_k^2}+2n\,$. By AM-GM:
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. |
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