Question

For positive reals a, b, c. Find the minimum value of $\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ba}$

• A. $2$
• B. $3$
• C. $4$
• D. None of these

Answer 1

The correct option is B $3$

We know if we have three real numbers $x,y$ and $z$ then

$AM\ge GM$ ........ $\left(1\right)$

i.e $\frac{x+y+z}{3}\ge \sqrt[3]{3xyz}$

Here the three positive numbers are $\frac{a^2}{bc}$ , $\frac{b^2}{ac}$ and $\frac{c^2}{ab}$

Applying inequality $\left(1\right)$, we get

$\frac{\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}}{3}\ge \sqrt[3]{3\frac{a^2}{bc}\times \frac{b^2}{ac}\times \frac{c^2}{ab}}$

$\Rightarrow$ $\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}\ge 3$

So the minimum value of $\frac{a^2}{bc}$ $+\frac{b^2}{ac}+\frac{c^2}{ab}$ is $3$

Hence, option B is correct.