Difference between revisions of "1977 USAMO Problems/Problem 5"

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== Solution ==
 
== Solution ==
By applying the Cauchy-Schwarz Inequality in the form <math>((\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</math>, we can easily reduce the given inequality to <math>0 \le 6\left(\sqrt{\frac {p}{q}} - \sqrt {\frac{q}{p}}\right)^2</math>, which is true by the Trivial Inequality. We see that equality is achieved when <math>\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}}</math>, which is achieved when <math>p=q</math>.
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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.
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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.
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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.
  
 
== See Also ==
 
== See Also ==

Revision as of 10:41, 18 June 2015

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.

See Also

1977 USAMO (ProblemsResources)
Preceded by
Problem 4
Followed by
Last Question
1 2 3 4 5
All USAMO Problems and Solutions

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