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.
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