Indian Math Olympiad 2002 Q3 | Can you solve the inequality?

Welcome! In this video, we will be going through problem 3 from the Indian Math Olympiad 2002. Hope you enjoy the video, and happy problem solving!

00:00 - Intro
00:08 - The Question
00:38 - Hints
01:28 - Solution
04:45 - Thanks for watching!

[gigagrzybiarz]

That was really easy for an Olympiad Problem

[michaelgreenberg]

Also, [a+a+a+(4-3a)]/4 >= [a^3(4-3a)]^(1/4) by the AM-GM inequality. This is equivalent to a^3(4-3a)<=1 (so you can cleverly avoid a lot of the algebra).

[harshvadher]

Actually you know that you solve by am-gm and get initially x³y³≤1 and then go for x³+y³=t then let 2³=t+3x²y+3xy² and again go for am-gm by having denominator=n=3. Thus u get same result but slightly greater as it says
X³y³(x³+y³)≤2.1

[mrhatman]

I don t know if this is a right solution but after aplying the gm-am inequality and finding xy≤1 we can find (xy)^3≤1 thus if we want to prove 2(xy)^3(4-3xy)≤2 we devide by both sides with 2 which means we have to prove (xy)^3×(4-3xy)≤1 but (xy)^3≤1 so we have to prove 4-3xy≤1 which means we have to prove xy≥1 but we proved xy≤1 so xy=1putting this value for xy into the inequality we get that it satifies it

[karthikkrishnaswami]

Very easy for an Olympiad problem. When we rewrite in terms of a, we could induct also isn't it? Just thinking....