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 $x = h$ 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 $17 + 4(\frac{h}{k} + \frac{k}{h})$ . Similarly if four are $k$ and one is $h$ . If three are $h$ and two are $k$ , then the LHS is $13 + 6(\frac{h}{k} + \frac{k}{h})$ . Similarly if three are $k$ and two are $h$ .

$\frac{h}{k} + \frac{k}{h} \geq 2$ 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 $13 + 6(\frac{h}{k} + \frac{k}{h})$ , 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.

1977 USAMO (Problems • Resources)
Preceded by Problem 4	Followed by Last Question
1 • 2 • 3 • 4 • 5
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1977 USAMO Problems/Problem 5

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