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If $\left\{b_{1}, \ldots, b_{n}\right\}$ is a permutation of the sequence $\left\{a_{1}, \ldots, a_{n}\right\}$ of positive reals, then prove that
\frac{a_{1}}{b_{1}}+\frac{a_{2}}{b_{2}}+\cdots+\frac{a_{n}}{b_{n}} \geq n
in Mathematics by Diamond (78,454 points) | 421 views

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Notice that $a_{1} a_{2} \cdots a_{n}=b_{1} b_{2} \cdots b_{n}$ since $b$ is a permutation of $a$. Then by AM-GM inequality
\frac{a_{1}}{b_{1}}+\frac{a_{2}}{b_{2}}+\cdots+\frac{a_{n}}{b_{n}} \geq n \sqrt[n]{\frac{a_{1}}{b_{1}} \frac{a_{2}}{b_{2}} \cdots \frac{a_{n}}{b_{n}}}=n \sqrt[n]{\frac{a_{1} a_{2} \cdots a_{n}}{a_{1} a_{2} \cdots a_{n}}}=n
by Diamond (78,454 points)

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