Question

Find minimum value $4x+\frac{9}{x}x>0,$

Answer 1

As we know that

$AM\ge GM$

$\frac{4x+\frac{9}{x}}{2}\ge \sqrt{\left(4x\right)(\frac{9}{x})}$

$\frac{4x+\frac{9}{x}}{2}\ge \sqrt{36}$

$4x+\frac{9}{x}\ge 12$

The minimum value of $4x+\frac{9}{x}$ is $12$