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Difference between revisions of "1977 USAMO Problems/Problem 5"
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== Solution ==
 
== Solution ==
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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 <math>x</math>. Then the expression on the LHS has the form <math>(r + x)(s + \frac{1}{x}) = (rs + 1) + sx + \frac{r}{x}</math>, where <math>r</math> and <math>s</math> are fixed. But this is convex. That is to say, as <math>x</math> increases if first decreases, then increases. So its maximum must occur at <math>x = p</math> or <math>x = q</math>. This is true for each variable.
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Suppose all five are <math>p</math> or all five are <math>q</math>, then the LHS is 25, so the inequality is true and strict unless <math>p = q</math>. If four are <math>p</math> and one is <math>q</math>, then the LHS is <math>17 + 4\left(\frac{p}{q} + \frac{q}{p}\right)</math>. Similarly if four are <math>q</math> and one is <math>p</math>. If three are <math>p</math> and two are <math>q</math>, then the LHS is <math>13 + 6\left(\frac{p}{q} + \frac{q}{p}\right)</math>. Similarly if three are <math>q</math> and two are <math>p</math>.
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<math>\frac{p}{q} + \frac{q}{p} \geq 2</math> with equality iff <math>p = q</math>, so if <math>p < q</math>, 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 <math>13 + 6\left(\frac{p}{q} + \frac{q}{p}\right)</math>, so the inequality is true with equality iff either (1) <math>p = q</math> or (2) three of <math>v, w, x, y, z</math> are <math>p</math> and two are <math>q</math> or vice versa.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 22:08, 27 June 2018

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 by
Problem 4 		Followed by
Last Question
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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