Difference between revisions of "2012 USAJMO Problems/Problem 3"
(→See Also) |
(→Solution 2) |
||
Line 27: | Line 27: | ||
<cmath>\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{1 + 3}{6} (a^2 + b^2 + c^2) = \frac{2}{3} (a^2 + b^2 + c^2).</cmath> | <cmath>\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{1 + 3}{6} (a^2 + b^2 + c^2) = \frac{2}{3} (a^2 + b^2 + c^2).</cmath> | ||
==Solution 2== | ==Solution 2== | ||
− | Split up the numerators and multiply both sides of the fraction by either a or 3a | + | Titu's Lemma: The sum of multiple fractions in the form <math>\frac{a_n^2}{b_n}</math> where <math>a_n</math> and <math>b_n</math> are sequences of real numbers is greater than of equal to the square of the sum of all <math>a_i</math> divided by the sum of all <math>b_i</math>, where i is a whole number less than n+1. Titu's Lemma can be proved using the Cauchy-Schwarz Inequality after multiplying out the denominator of the RHS. |
− | < | + | |
− | Then use Titu's Lemma: < | + | Consider the LHS of the proposed inequality. Split up the numerators and multiply both sides of the fraction by either a or 3a to make the LHS |
− | + | <cmath>\sum_{cyc} \frac {a^4} {5a^2+ab}+\sum_{cyc} \frac {9a^4} {15ac+3a^2}</cmath> | |
− | + | (Cyclic notation is a special notation in which all permutations of (a,b,c) is the summation command.) | |
+ | |||
+ | Then use Titu's Lemma on all terms: <cmath>\sum_{cyc} \frac {a^4} {5a^2+ab} +\sum_{cyc} \frac {9a^4} {15ac+3a^2} \ge \frac {16(a^2+b^2+c^2)^2} {8(a^2+b^2+c^2)+16(ab+bc+ca)} \ge \frac {16(a^2 + b^2 + c^2)^2}{24(a^2 + b^2 + c^2)} = \frac{2}{3} (a^2 + b^2 + c^2) </cmath> owing to the fact that <math>a^2+b^2+c^2 \ge ab+bc+ca</math>, which is actually equivalent to <math>(a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0</math>! | ||
==Solution 3== | ==Solution 3== |
Revision as of 19:48, 12 April 2014
Problem
Let , , be positive real numbers. Prove that
Solution
By the Cauchy-Schwarz inequality, so Since , Hence,
Again by the Cauchy-Schwarz inequality, so Since , Hence,
Therefore,
Solution 2
Titu's Lemma: The sum of multiple fractions in the form where and are sequences of real numbers is greater than of equal to the square of the sum of all divided by the sum of all , where i is a whole number less than n+1. Titu's Lemma can be proved using the Cauchy-Schwarz Inequality after multiplying out the denominator of the RHS.
Consider the LHS of the proposed inequality. Split up the numerators and multiply both sides of the fraction by either a or 3a to make the LHS (Cyclic notation is a special notation in which all permutations of (a,b,c) is the summation command.)
Then use Titu's Lemma on all terms: owing to the fact that , which is actually equivalent to !
Solution 3
We proceed to prove that
(then the inequality in question is just the cyclic sum of both sides, since )
Indeed, by AP-GP, we have
and
Summing up, we have
which is equivalent to:
Dividing from both sides, the desired inequality is proved.
--Lightest 15:31, 7 May 2012 (EDT)
See Also
2012 USAJMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.