In part 1, we saw what Titu's Lemma is and one example of using it. Now, let's see more examples of using it.
(TOT 1998) Prove that for any positive real numbers ,
Again, we want to make the numerators on the left hand side be squares.
So, we again multiply the numerators and denominators of the fractions by respectively.
By Titu's Lemma,
But the denominator is equivalent to .
Thus, we the inequality becomes
Finally, we have to prove that , which is true. (The proof is similar to the last problem.)
(IMO 1995) Let be positive real numbers such that . Prove that
Solution : Direct application of Titu's Lemma is unfortunately doomed to failure. We divide the numerator and denominator by and respectively.
Now, we have by Titu's Lemma,
where the last inequality holds by AM-GM.
This proves the inequality.
Prove that for all positive real numbers
Prove that for all positive real numbers such that , then
Note : Problem 2 uses the same idea as Example 2, but Problem 4 is a bit harder, for a few reasons :
There are no square terms on the numerators.
The inequality sign is less than or equal to.