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

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== Solution ==
 
== Solution ==
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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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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 = h</math> or <math>x = k</math>. 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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Suppose all five are <math>h</math> or all five are <math>k</math>, then the LHS is 25, so the inequality is true and strict unless <math>h = k</math>. If four are <math>h</math> and one is <math>k</math>, then the LHS is <math>17 + 4(\frac{h}{k} + \frac{k}{h})</math>. Similarly if four are <math>k</math> and one is <math>h</math>. If three are <math>h</math> and two are <math>k</math>, then the LHS is <math>13 + 6(\frac{h}{k} + \frac{k}{h})</math>. Similarly if three are <math>k</math> and two are <math>h</math>.
  
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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.
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<math>\frac{h}{k} + \frac{k}{h} \geq 2</math> with equality iff <math>h = k</math>, so if <math>h < k</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(\frac{h}{k} + \frac{k}{h})</math>, so the inequality is true with equality iff either (1) <math>h = k</math> or (2) three of <math>v, w, x, y, z</math> are <math>h</math> and two are <math>k</math> or vice versa.
  
 
== See Also ==
 
== See Also ==

Revision as of 14:30, 19 November 2017

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.

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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