Difference between revisions of "1977 USAMO Problems/Problem 5"
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== Solution == | == Solution == | ||
| − | {{ | + | 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 == | ||
{{USAMO box|year=1977|num-b=4|after=Last Question}} | {{USAMO box|year=1977|num-b=4|after=Last Question}} | ||
| + | {{MAA Notice}} | ||
[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
[[Category:Olympiad Inequality Problems]] | [[Category:Olympiad Inequality Problems]] | ||
Latest revision as of 23:08, 27 June 2018
Problem
If 
are positive numbers bounded by 
and 
, i.e, if they lie in ![$[p,q], 0 < p$](http://latex.artofproblemsolving.com/b/7/d/b7db0d1c44b17ab521242db4599465126d5ad617.png)
, prove that
![\[(a+b +c +d +e)\left(\frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e}\right) \le 25 + 6\left(\sqrt{\frac {p}{q}} - \sqrt {\frac{q}{p}}\right)^2\]](http://latex.artofproblemsolving.com/7/d/4/7d42a948da6239cf2d21f08336d9b2ad4f8a65b0.png)
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 
. Then the expression on the LHS has the form 
, where 
and 
are fixed. But this is convex. That is to say, as increases if first decreases, then increases. So its maximum must occur at 
or 
. This is true for each variable.
Suppose all five are or all five are , then the LHS is 25, so the inequality is true and strict unless 
. If four are and one is , then the LHS is 
. Similarly if four are and one is . If three are and two are , then the LHS is 
. Similarly if three are and two are .
with equality iff , so if 
, 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 , so the inequality is true with equality iff either (1) or (2) three of 
are and two are 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
