Difference between revisions of "2001 USAMO Problems/Problem 3"
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<cmath>ab + bc + ca - abc = a(b + c) + bc(1-a) \ge 0.</cmath> | <cmath>ab + bc + ca - abc = a(b + c) + bc(1-a) \ge 0.</cmath> | ||
− | + | Now, without loss of generality, we assume that <math>b</math> and <math>c</math> are either both greater than 1 or both less than one, so <math>(b-1)(c-1)\ge 0</math>. From the given equation, we can express <math>a</math> in terms of <math>b</math> and <math>c</math> as | |
<center> <math>a=\frac{\sqrt{(4-b^2)(4-c^2)}-bc}{2} </math></center> | <center> <math>a=\frac{\sqrt{(4-b^2)(4-c^2)}-bc}{2} </math></center> | ||
Thus, | Thus, |
Revision as of 17:50, 31 May 2011
Problem
Let and satisfy

Show that

Solution
First we prove the lower bound.
Note that we cannot have all greater than 1.
Therefore, suppose
.
Then
Now, without loss of generality, we assume that and
are either both greater than 1 or both less than one, so
. From the given equation, we can express
in terms of
and
as

Thus,

From Cauchy,

This completes the proof.
See also
2001 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |