# 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 + 1/x) = (rs + 1) + sx + 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 = h or x = k. This is true for each variable.

Suppose all five are h or all five are k, then the lhs is 25, so the inequality is true and strict unless h = k. If four are h and one is k, then the lhs is 17 + 4(h/k + k/h). Similarly if four are k and one is h. If three are h and two are k, then the lhs is 13 + 6(h/k + k/h). Similarly if three are k and two are h.

h/k + k/h ≥ 2 with equality iff h = k, so if h < k, 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(h/k + k/h), so the inequality is true with equality iff either (1) h = k or (2) three of v, w, x, y z are h and two are k or vice versa.