If $a^{2} b^{3} c = \frac{256}{27}$ find min value of a + b + c, given a, b, c are positive real nos.

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Solution

The correct option is D 4
Take numbers as $\frac{a}{2}, \frac{a}{2}, \frac{b}{3}, \frac{b}{3}, \frac{b}{3}, c$

Now apply AM $>$ GM on these six numbers.

We have $a^{2} b^{3} c = \frac{256}{27}$ . Solving we get min value of a+b+c as 4.

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a, b, c

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a, b, c

\frac{a + b - c}{c} = \frac{a - b + c}{b} = \frac{- a + b + c}{a},

\frac{(a + b) (b + c) (c + a)}{a b c} .

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