2012 USAJMO Problems/Problem 3
Let , , be positive real numbers. Prove that
By the Cauchy-Schwarz inequality, so Since , Hence,
Again by the Cauchy-Schwarz inequality, so Since , Hence,
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 !
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
Summing up, we have
which is equivalent to:
Dividing from both sides, the desired inequality is proved.
--Lightest 15:31, 7 May 2012 (EDT)
(by AM-GM) as desired.
Lemma: For all positive integers , we have . Proof: We want to prove . We can factor a from this to get . The 2nd factor of the LHS is equal to , which is positive since . Additionally, by the Trivial Inequality, is positive, so the LHS is positive, proving the lemma.
We now go back to the equation. We have by the lemma, and the RHS equals . Therefore, we are done.
--jeteagle 2:18, 31 December 2019 (EDT)
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