Why is it always minimum?

Algebra Level 3

For positive real numbers a , b , c a, b, c , find the minimum integer value possible of the following equation:

6 a 3 + 9 b 3 + 32 c 3 + 1 4 a b c 6a^{3} + 9b^{3} + 32c^{3} + \frac{1}{4abc}

Hint: Click here for hint.


The answer is 6.

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1 solution

Mohammad Al Ali
Jun 1, 2014

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 = 32 c 3 x_1 = 6a^{3} , x_2 =9b^{3}, x_3 = 32c^{3} we obtain,

6 a 3 + 9 b 3 + 32 c 3 3 1728 a 3 b 3 c 3 3 \frac{6a^{3}+9b^{3}+32c^{3}}{3} \ge \sqrt[3]{1728a^{3}b^{3}c^{3}}

Which simplifies nicely into,

6 a 3 + 9 b 3 + 32 c 3 36 a b c 6a^{3}+9b^{3}+32c^{3} \ge 36abc

Thus, adding 1 4 a b c \frac{1}{4abc} on both sides we get,

6 a 3 + 9 b 3 + 32 c 3 + 1 4 a b c 36 a b c + 1 4 a b c 6a^{3}+9b^{3}+32c^{3} + \frac{1}{4abc} \ge 36abc + \frac{1}{4abc}


Applying AM-GM on RHS on the inequality, x 1 = 36 a b c , x 2 = 1 4 a b c x_1 = 36abc, x_2 = \frac{1}{4abc} we attain,

36 a b c + 1 4 a b c 2 36 a b c × 1 4 a b c \frac{36abc + \frac{1}{4abc}}{2} \ge \sqrt{36abc \times \frac{1}{4abc}}

36 a b c + 1 4 a b c 6 36abc + \frac{1}{4abc} \ge 6

Thus,

6 a 3 + 9 b 3 + 32 c 3 + 1 4 a b c 36 a b c + 1 4 a b c 6 6a^{3}+9b^{3}+32c^{3} + \frac{1}{4abc} \ge 36abc + \frac{1}{4abc} \ge 6

Minimum value is 6.

Alternatively, we could have splited up 1 4 a b c \frac {1}{4abc} into 1 12 a b c + 1 12 a b c + 1 12 a b c \frac {1}{12abc}+\frac {1}{12abc}+\frac {1}{12abc} and then use AM-GM.

Xuming Liang - 7 years ago

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Yep, that's how I did it. :)

Pranav Arora - 7 years ago

What will you do after splitting it in this way??? I solved it the way Finn and Mohammad did it

Tanya Gupta - 7 years ago

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Isn't that obvious? Apply AM-GM inequality. :P

Pranav Arora - 7 years ago

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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!!!

Tanya Gupta - 7 years ago

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@Tanya Gupta Hahah its trivial

Magnas Bera - 1 year, 11 months ago

The proof is not correct because the fact that the value 6 6 can be achieved has not been proven. It actually can, though, and the equality holds when a = 1 6 3 a=\frac{1}{\sqrt[3]{6}} , b = 1 9 3 b=\frac{1}{\sqrt[3]{9}} and c = 1 32 3 c=\frac{1}{\sqrt[3]{32}} .

mathh mathh - 6 years, 12 months ago

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

Finn Hulse - 7 years ago

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