Track / Stage 6 / 24 of 400 #1024 of 1964

Problem 1024

National olympiad, first round

Algebra Difficulty 6.0 Prove it

Let $n$ be a strictly positive integer, $e_{1}, \cdots, e_{n}$ any real numbers, and $f_{1}, \cdots, f_{n}$ strictly positive real numbers. Show that:

$\frac{e_{1}^{2}}{f_{1}}+\cdots+\frac{e_{n}^{2}}{f_{n}} \geqslant \frac{\left(e_{1}+\cdots+e_{n}\right)^{2}}{f_{1}+\cdots+f_{n}}$

Determine the cases of equality.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Apply the Cauchy-Schwarz inequality with $a_{i}=\frac{e_{i}}{\sqrt{f_{i}}}$ and $b_{i}=\sqrt{f_{i}}$ . We then have

$\left(f_{1}+f_{2}+\ldots+f_{n}\right)\left(\frac{e_{1}^{2}}{f_{1}}+\ldots+\frac{e_{n}^{2}}{f_{n}}\right) \geq\left(e_{1}+e_{2}+\ldots+e_{n}\right)^{2}$

and the result follows by dividing by the sum of the $f_{i}$ . Equality holds when the $a_{i}$ are proportional to the $b_{i}$ . This is equivalent to saying that all the $e_{i}$ are zero or that there exists a real number $\lambda$ such that for all $i$ ,

$\sqrt{f_{i}}=\lambda \frac{e_{i}}{\sqrt{f_{i}}}$

or that the $e_{i}$ and $f_{i}$ are proportional.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.