After a long time

Algebra Level 3

Let $a,b,c$ be positive reals satisfying $abc = 3$ . Find the minimum value of $a^{2} b + b^{2} c + c^{2} a$ .

The answer is 9.

1 solution

Chew-Seong Cheong
Nov 24, 2015

Since $a,b,c >0$ , we can apply AM-GM Inequality.

$\Rightarrow a^2b+b^2c+c^2a \ge 3 \sqrt[3]{a^3b^3c^3} = 3abc = \boxed{9}$

Did the same way sir.. AM-GM inequality is simple, yet elegant

Jun Arro Estrella - 5 years, 5 months ago

I am glad you like it.

Chew-Seong Cheong - 5 years, 5 months ago

can't we take values satisfy d eq ... won't we get sol...

Hemanth Sparks - 5 years, 6 months ago

There are 3 unknown a, b and c but only 1 equation abc=3. There are infinite solutions but only one minimum.

Actually, since the equation $abc = 9$ and $a^2b+b^2c+c^2a$ and symmetrical, we can assume $a=b=c=\sqrt[3]{3}$ for minimum $a^2b+b^2c+c^2a = 3+3+3 = \boxed{9}$ .

Chew-Seong Cheong - 5 years, 6 months ago

After a long time

The answer is 9.

1 solution

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