Difference between revisions of "2012 USAJMO Problems/Problem 3"
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It suffices to prove that<math> \frac {16(a^2+b^2+c^2)^2} {8(a^2+b^2+c^2)+16(ab+bc+ca)} \ge \frac {2} {3} (a^2+b^2+c^2)</math> | It suffices to prove that<math> \frac {16(a^2+b^2+c^2)^2} {8(a^2+b^2+c^2)+16(ab+bc+ca)} \ge \frac {2} {3} (a^2+b^2+c^2)</math> | ||
After some simplifying, it reduces to <math>a^2+b^2+c^2 \ge ab+bc+ca</math> which is trivial by the Rearrangement Inequality. -r31415 | After some simplifying, it reduces to <math>a^2+b^2+c^2 \ge ab+bc+ca</math> which is trivial by the Rearrangement Inequality. -r31415 | ||
+ | |||
+ | ==Solution 3== | ||
+ | We proceed to prove that | ||
+ | <cmath>\frac{a^3 + 3b^3}{5a + b} \ge -\frac{1}{36} a^2 + \frac{25}{36} b^2</cmath> | ||
+ | |||
+ | Indeed, by AP-GP, we have | ||
+ | |||
+ | <cmath> 41 (a^3 + b^3+b^3) \ge 41*3 ab^2 </cmath> | ||
+ | |||
+ | and | ||
+ | |||
+ | <cmath> b^3 + a^2b \ge 2 ab^2</cmath> | ||
+ | |||
+ | Summing up, we have | ||
+ | |||
+ | <cmath> 41a^3 + 83b^3 + a^2 b \ge 125 ab^2</cmath> | ||
+ | |||
+ | which is equivalent to: | ||
+ | |||
+ | <cmath> 36(a^3 + 23b^3) \ge (5a + b)(-a^2 + 25b^2) </cmath> | ||
+ | |||
+ | Dividing <math>36(5a+b)</math> from both sides, the desired inequality is proved. | ||
+ | |||
+ | Permuting the variables, we have all these three inequalities: | ||
+ | |||
+ | <cmath>\frac{a^3 + 3b^3}{5a + b} \ge -\frac{1}{36} a^2 + \frac{25}{36} b^2</cmath> | ||
+ | |||
+ | <cmath>\frac{b^3 + 3c^3}{5b + c} \ge -\frac{1}{36} b^2 + \frac{25}{36} c^2</cmath> | ||
+ | |||
+ | <cmath>\frac{c^3 + 3a^3}{5c + a} \ge -\frac{1}{36} c^2 + \frac{25}{36} a^2</cmath> | ||
+ | |||
+ | The inequality in question is just the sum of these, hence it is proved. | ||
==See Also== | ==See Also== |
Revision as of 15:30, 7 May 2012
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
Split up the numerators and multiply both sides of the fraction by either a or 3a: Then use Titu's Lemma: It suffices to prove that After some simplifying, it reduces to which is trivial by the Rearrangement Inequality. -r31415
Solution 3
We proceed to prove that
Indeed, by AP-GP, we have
and
Summing up, we have
which is equivalent to:
Dividing from both sides, the desired inequality is proved.
Permuting the variables, we have all these three inequalities:
The inequality in question is just the sum of these, hence it is proved.
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 |