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.

Suggest Corrections

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