Question

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

Answer 1

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

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 \(x\)