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

Fix four of the variables and allow the other to vary. Suppose, for example, we fix all but $x$. Then the expression on the LHS has the form $(r + x)(s + \frac{1}{x}) = (rs + 1) + sx + \frac{r}{x}$, where $r$ and $s$ are fixed. But this is convex. That is to say, as $x$ increases if first decreases, then increases. So its maximum must occur at $x = p$ or $x = q$. This is true for each variable.

Suppose all five are $p$ or all five are $q$, then the LHS is 25, so the inequality is true and strict unless $p = q$. If four are $p$ and one is $q$, then the LHS is $17 + 4\left(\frac{p}{q} + \frac{q}{p}\right)$. Similarly if four are $q$ and one is $p$. If three are $p$ and two are $q$, then the LHS is $13 + 6\left(\frac{p}{q} + \frac{q}{p}\right)$. Similarly if three are $q$ and two are $p$.

$\frac{p}{q} + \frac{q}{p} \geq 2$ with equality iff $p = q$, so if $p < q$, then three of one and two of the other gives a larger LHS than four of one and one of the other. Finally, we note that the RHS is in fact $13 + 6\left(\frac{p}{q} + \frac{q}{p}\right)$, so the inequality is true with equality iff either (1) $p = q$ or (2) three of $v, w, x, y, z$ are $p$ and two are $q$ or vice versa.

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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