2003 USAMO Problems/Problem 5
Contents
Problem
Let , , be positive real numbers. Prove that
Solution
Solution 1
Since all terms are homogeneous, we may assume WLOG that .
Then the LHS becomes .
Notice , so .
So , as desired.
Solution 2
Note that Setting and yields Thus, we have
\[\frac{(2a + b + c)^2}{2a^2 + (b + c)^2 = \frac{3(2a^2 + (b + c)^2) - 2(a - b - c)^2}{2a^2 + (b + c)^2} = 3 - \frac{2(a - b - c)^2}{2a^2 + (b + c)^2},\] (Error compiling LaTeX. ! File ended while scanning use of \frac .)
and its analogous forms. Thus, the desired inequality is equivalent to Because , we have and its analogous forms. It suffices to show that or, Multiplying this out the left-hand side of the last inequality gives . Therefore the inequality is equivalent to , which is evident because Equality holds when .
Solution 3
Given a function of three variables, define the cyclic sum We first convert the inequality into Splitting the 5 among the three terms yields the equivalent form The numerator of the term shown factors as , where . We will show that Indeed, is equivalent to which reduces to evident. We proved that hence follows. Equality holds if and only if , i.e., when .
Solution 4
Given a function of variables, we define the symmetric sum where runs over all permutations of (for a total of terms).
We combine the terms in the desired inequality over a common denominator and use symmetric sum notation to simplify the algebra. The numerator of the difference between the two sides is Recalling Schur's Inequality, we have or Hence, and by repeated AM-GM inequality, and Adding these three inequalities yields the desired result.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
2003 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |