Prove the product of sums of n numbers and their reciprocals is greater than n^2

Let $x_1, \ldots, x_n \in \mathbb{R}_{>0}$ and let $y_k = \frac{1}{x_k}$ for $k = 1, \ldots, n$ . Prove

$\left( \sum_{k=1}^n x_k \right) \left( \sum_{k=1}^n y_k \right) \geq n^2.$

Proof. First,

$y_k = \frac{1}{x_k} \quad \implies \quad \sqrt{y_k} = \frac{1}{\sqrt{x_k}} \quad \implies \quad \sqrt{x_k} \sqrt{y_k} = 1$

for all $k = 1, \ldots, n$ . We then apply the Cauchy-Schwarz inequality to the numbers $\sqrt{x_1}, \ldots, \sqrt{x_n}$ and $\sqrt{y_1}, \ldots, \sqrt{y_n}$ . So, Cauchy-Schwarz gives us,

$\begin{align*} &&\left( \sum_{k=1}^n a_k b_k \right)^2 &\leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)\\ \implies && \left( \sum_{k=1}^n \sqrt{x_k} \sqrt{y_k} \right)^2 &\leq \left( \sum_{k=1}^n (\sqrt{x_k})^2 \right) \left( \sum_{k=1}^n (\sqrt{y_k})^2 \right) \\ \implies && \left( \sum_{k=1}^n 1 \right)^2 &\leq \left( \sum_{k=1}^n x_k \right) \left( \sum_{k=1}^n y_k \right)\\ \implies && n^2 &\leq \left(\sum_{k=1}^n x_k \right) \left( \sum_{k=1}^n y_k \right). \qquad \blacksquare \end{align*}$

One comment

September 7, 2019 at 10:15 pm

richardsull says:

You can also prove it this way:

If $M_1(a_k)$ and $G(a_k)$ are the arithemetic and geometric means of the sequence $a_1, a_2, \ldots, a_n$ , then

$\begin{align*} \left(\sum_{k=1}^{n}{x_k}\right)\left(\sum_{k=1}^{n}{y_k}\right) &= n^2\left(\frac{1}{n}\sum_{k=1}^{n}{x_k}\right)\left(\frac{1}{n}\sum_{k=1}^{n}{y_k}\right)\\ &= n^2 M_1(x_k)\cdot M_1(y_k)\\ &\ge n^2 G(x_k)\cdot G(y_k)\\ &= n^2 G(x_k y_k)\\ &= n^2 \cdot 1\\ &= n^2 \end{align*}$

Stumbling Robot

Prove the product of sums of n numbers and their reciprocals is greater than n^2

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