\sqrt{\frac{\left(2x+2\right)}{x+2}}-\sqrt{\frac{\left(x+2\right)}{2x+2}}\ge \:\frac{7}{12}

Question

imqwerty · Answer

$\Large\sqrt{\frac{\left(2x+2\right)}{x+2}}-\sqrt{\frac{\left(x+2\right)}{2x+2}}\ge \:\frac{7}{12}$

Lets say that   $\Large\sqrt{\frac{\left(2x+2\right)}{x+2}}=\alpha$ 

so we have this->

$\Large \alpha - \frac{1}{\alpha} \ge \frac{7}{12}$ 

solve this and find range of $\alpha$

$\alpha$ is nothing but this-> $\large \sqrt{\frac{\left(2x+2\right)}{x+2}}$ 

so you have the range of $\large \sqrt{\frac{\left(2x+2\right)}{x+2}}$ you can find range of $x$