Let a,b,c be positive real numbers and$a{x}^2+\frac{b}{x^2}\ge c$ all $x\in R^{+}.$ Then

Let $f\left(x\right)=a{x}^2+\frac{b}{x^2}-c$ where $a,b,c>0\text{ and }x>0$

Then $f^{\mathrm{\prime }}\left(x\right)=2ax-\frac{2b}{x^3}\text{ and }{f}^{\prime \prime }\left(x\right)=2a+\frac{6b}{x^4}$

For local maximum or minimum, we must have

$f^{\mathrm{\prime }}\left(x\right)=0\Rightarrow x^4=\frac{b}{a}\Rightarrow x=\pm {\left(\frac{b}{a}\right)}^{1/4}$

Clearly, $f^{\ast }\left\{\pm {\left(\frac{b}{a}\right)}^{1/+}\right\}=2a+6a>0$

Therefore $x=\pm {\left(\frac{b}{a}\right)}^{1/4}$ are points of local minimum

Local minimum value of f (x) is given by

$f\left\{{\left(\frac{b}{a}\right)}^{1/4}\right\}=a{\left(\frac{b}{a}\right)}^{1/2}+b{\left(\frac{a}{b}\right)}^{1/2}-c=2\sqrt{ab}-c$

But, it is given that

$\begin{array}{c}f\left(x\right)\ge 0\text{ for all }x\\ \therefore 2\sqrt{ab}-c\ge 0\Rightarrow 4ab\ge c^2\end{array}$