Question

If $f\left(x\right)=x^2+\frac{1}{x^2}$, then find the minimum value of the function $g\left(x\right)=f\left(x\right)+f\left(\frac{1}{x}\right).$

Answer 1

$g\left(x\right)=(x^2+(\frac{1}{x^2}\left)\right)+\left(\right(\frac{1}{x}{)}^2+\left(\frac{1}{(\frac{1}{x}{)}^2}\right))$

$=2(x^2+(\frac{1}{x^2}\left)\right)nowfor(x^2+(\frac{1}{x^2}),applyAM⩾GM$

$\left(\frac{(x^2+(\frac{1}{x^2})}{2}\right)⩾\sqrt{(x^2\times (\frac{1}{x^2})}\therefore (x^2+(\frac{1}{x^2})⩾2$

$minvalueof(x^2+(\frac{1}{x^2})=2$

$andhenceminvalueof\left(x\right)=2\times 2=4$