We explorean IMO Inequality problem using titu's lemma which is.an important tool in olympiad inequality as well as jee mains, advanced and ISI/CMI
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imo inequality, olympiad, three line solution, ans academy, math olympiad, inequality tricks, math olympiad algebra problem, Jee main jee advanced, Jee advanced, inequality methods, imo tricks, math, math olympiad preparation, olympiad math, algebra solutions
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00:00Okay guys, so we head up to another example. This example appeared in the IMO problem.
00:04So in the IMO 1995, this problem was the second one. So this inequality, as you can see, 1
00:10upon a cube into b plus c plus 1 upon b cube into c plus a plus 1 upon c cube into a plus
00:14t. We needed to prove that this is greater than equal to 3 by 2. There was an additional
00:19condition that a, b, c are all positive reals and a, b, c equals to 1. Okay. So we needed
00:26to prove this. Now, we again applied T2's lemma to prove this one. T2's lemma is so
00:32much handworthy that you can apply it almost everywhere and this problem can be solved
00:37by this. We can arrange, we take only the left hand side expression and call it S. So
00:43we need to prove actually S is greater than equal to 3 by 2. So S, we arrange it like
00:48this. So we arrange it like 1 upon a square and keep the rest of the a and multiply it
01:06out with them. Same for all of them 3, like this. Now you can see that the numerator is
01:11in the T2's lemma form. So we can easily say that 1 upon a square is 1 upon a whole square
01:16and all of them are in the T2's lemma for the case 3. This will, on applying T2's lemma will
01:21give you 1 upon a plus 1 upon b plus 1 upon c whole square and down there it will be 2 times
01:31a, b plus b, c plus c, a. Okay. Now this a, b, b, c, c, a, if you multiply it by c up and down,
01:43a up and down and b up and down, you can see by employing a, b, c equals to 1, you can write this
01:47as this on the numerator and this on the denominator. Not this actually, like this. Okay. Now this to
02:05cut off like this. So we are left only with this one. Now we can apply a and gm to this
02:11quantities, 1 upon a, 1 upon b and 1 upon c to get this half comes out and there is a gm which
02:17means 3 times cube root of 1 upon a, b, c. 1 upon a into 1 upon b into 1 upon c. Now this a,
02:27b, c is 1. You can put it there and find it is only 3 by 2. So we have just proved this in 3 lines.
02:34We have proved it in 3 lines using T2's lemma. What you could have done that you could have
02:39substituted 1 upon a, 1 upon b, 1 upon c as x, y and z. And then after this, the same method could
02:45have been applied, but this, this was also the same method, but T2's lemma is so much handworthy
02:49that it did the ammo problem in just 3 steps. So this is the ammo problem and I hope do not be,
02:55do not be helpful by saying that this is a short one because it's actually a lengthy one,
03:01but this problem didn't appear and there were no solutions using T2's lemma. So thanks for
03:05watching. So we'll go to the next example.