Difference between revisions of "2000 USAMO Problems/Problem 6"
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Each term in the summation is non-negative, so the sum itself is non-negative. <math>\blacksquare</math> | Each term in the summation is non-negative, so the sum itself is non-negative. <math>\blacksquare</math> | ||
− | We now define <math>r_i=\frac{\max(a_i,b_i)}{\min(a_i,b_i)}-1</math>. If <math>\min(a_i,b_i)=0</math>, then let <math>r_i</math> be any non-negative number. Define <math>x_i=\sgn(a_i-b_i)\min(a_i,b_i)</math>. | + | We now define <math>r_i=\frac{\max(a_i,b_i)}{\min(a_i,b_i)}-1</math>. If <math>\min(a_i,b_i)=0</math>, then let <math>r_i</math> be any non-negative number. Define <math>x_i=\text{sgn}(a_i-b_i)\min(a_i,b_i)</math>. |
'''Lemma 2:''' <math>\min(a_{i}b_{j}, a_{j}b_{i})-\min(a_{i}a_{j}, b_{i}b_{j}) =\min(r_{i}, r_{j}) x_{i}x_{j}</math> | '''Lemma 2:''' <math>\min(a_{i}b_{j}, a_{j}b_{i})-\min(a_{i}a_{j}, b_{i}b_{j}) =\min(r_{i}, r_{j}) x_{i}x_{j}</math> |
Latest revision as of 19:00, 10 March 2015
Problem
Let be nonnegative real numbers. Prove that
Solution
Credit for this solution goes to Ravi Boppana.
Lemma 1: If are non-negative reals and are reals, then
Proof: Without loss of generality assume that the sequence is increasing. For convenience, define . The LHS of our inequality becomes
This expression is equivalent to the sum
Each term in the summation is non-negative, so the sum itself is non-negative.
We now define . If , then let be any non-negative number. Define .
Lemma 2:
Proof: Switching the signs of and preserves inequality, so we may assume that . Similarly, we can assume that . If , then both sides are zero, so we may assume that and are positive. We then have from the definitions of and that
This means that
This concludes the proof of Lemma 2.
We can then apply Lemma 2 and Lemma 1 in order to get that
This implies the desired inequality.
See Also
2000 USAMO (Problems • Resources) | ||
Preceded by Problem 5 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.