Difference between revisions of "2012 USAJMO Problems/Problem 3"
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==Solution 3== | ==Solution 3== | ||
We proceed to prove that | 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> | + | <cmath>\frac{a^3 + 3b^3}{5a + b} \ge -\frac{1}{36} a^2 + \frac{25}{36} b^2 </cmath> |
− | (then the inequality in question is just the cyclic sum of both sides, since < | + | (then the inequality in question is just the cyclic sum of both sides, since <cmath>\sum_{cyc} (-\frac{1}{36} a^2 + \frac{25}{36} b^2) = \frac{24}{36}\sum_{cyc} a^2 = \frac{2}{3} (a^2+b^2+c^2)</cmath> |
+ | ) | ||
Indeed, by AP-GP, we have | Indeed, by AP-GP, we have | ||
− | <cmath> 41 (a^3 + b^3+b^3) \ge 41 | + | <cmath> 41 (a^3 + b^3+b^3) \ge 41 \cdot 3 ab^2 </cmath> |
and | and |
Revision as of 15:34, 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
(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 |