# 1977 USAMO Problems/Problem 5

## Problem

If $a,b,c,d,e$ are positive numbers bounded by $p$ and $q$, i.e, if they lie in $[p,q], 0 < p$, prove that $$(a+b +c +d +e)\left(\frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e}\right) \le 25 + 6\left(\sqrt{\frac {p}{q}} - \sqrt {\frac{q}{p}}\right)^2$$ and determine when there is equality.

## Solution

By applying the Cauchy-Schwarz Inequality in the form $((\sqrt{a})^2 +(\sqrt{b})^2 +(\sqrt{c})^2 +(\sqrt{d})^2 +(\sqrt{e})^2) \left( \left( \dfrac{1}{(\sqrt{a})^2} \right) + \left( \dfrac{1}{(\sqrt{b})^2} \right) + \left( \dfrac{1}{(\sqrt{c})^2} \right) + \left( \dfrac{1}{(\sqrt{d})^2} \right) + \left( \dfrac{1}{(\sqrt{e})^2} \right) \right) \le \left( \dfrac{\sqrt{a}}{\sqrt{a}} + \dfrac{\sqrt{b}}{\sqrt{b}} + \dfrac{\sqrt{c}}{\sqrt{c}} + \dfrac{\sqrt{d}}{\sqrt{d}} + \dfrac{\sqrt{e}}{\sqrt{e}} \right)^2 = (5)^2 = 25$, we can easily reduce the given inequality to $0 \le 6\left(\sqrt{\frac {p}{q}} - \sqrt {\frac{q}{p}}\right)^2$, which is true by the Trivial Inequality. We see that equality is achieved when $\left(\sqrt{\frac {p}{q}} - \sqrt {\frac{q}{p}}\right)^2 = 0 \rightarrow \left(\sqrt{\frac {p}{q}} - \sqrt {\frac{q}{p}}\right)=0 \rightarrow \sqrt{\frac {p}{q}}=\sqrt {\frac{q}{p}}$, which is achieved when $p=q$.

## See Also

 1977 USAMO (Problems • Resources) Preceded byProblem 4 Followed byLast Question 1 • 2 • 3 • 4 • 5 All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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