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 = | + | 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. |
− | Suppose all five are <math> | + | 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>. |
− | <math>\frac{ | + | <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 23:08, 27 June 2018
Problem
If are positive numbers bounded by
and
, i.e, if they lie in
, prove that
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.