For positive real numbers a , b , c , find the minimum integer value possible of the following equation:
6 a 3 + 9 b 3 + 3 2 c 3 + 4 a b c 1
Hint: Click here for hint.
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Alternatively, we could have splited up 4 a b c 1 into 1 2 a b c 1 + 1 2 a b c 1 + 1 2 a b c 1 and then use AM-GM.
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Yep, that's how I did it. :)
What will you do after splitting it in this way??? I solved it the way Finn and Mohammad did it
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Isn't that obvious? Apply AM-GM inequality. :P
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@Pranav Arora – OMG!!! I was assuming that he meant splitting it like this after doing the first step(getting 36abc)....I just realized that it is supposed to be the first step....Thanks!!!
The proof is not correct because the fact that the value 6 can be achieved has not been proven. It actually can, though, and the equality holds when a = 3 6 1 , b = 3 9 1 and c = 3 3 2 1 .
YES! So happy to have a solution utilizing AM-GM. I used literally the exact same method and I'm really happy because my knowledge of inequalities is very poor. Great problem! :D
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The main key in solving this equation is using the AM-GM Inequality.
Using the AM-GM Inequality with x 1 = 6 a 3 , x 2 = 9 b 3 , x 3 = 3 2 c 3 we obtain,
3 6 a 3 + 9 b 3 + 3 2 c 3 ≥ 3 1 7 2 8 a 3 b 3 c 3
Which simplifies nicely into,
6 a 3 + 9 b 3 + 3 2 c 3 ≥ 3 6 a b c
Thus, adding 4 a b c 1 on both sides we get,
6 a 3 + 9 b 3 + 3 2 c 3 + 4 a b c 1 ≥ 3 6 a b c + 4 a b c 1
Applying AM-GM on RHS on the inequality, x 1 = 3 6 a b c , x 2 = 4 a b c 1 we attain,
2 3 6 a b c + 4 a b c 1 ≥ 3 6 a b c × 4 a b c 1
3 6 a b c + 4 a b c 1 ≥ 6
Thus,
6 a 3 + 9 b 3 + 3 2 c 3 + 4 a b c 1 ≥ 3 6 a b c + 4 a b c 1 ≥ 6
Minimum value is 6.