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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*}

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