Question

For all real x, the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is

• A. $\frac{1}{3}$
• B. 1
• C. 3

Answer 1

The correct option is A $\frac{1}{3}$

Let $z=\frac{1-x+x^2}{1+x+x^2}$
$\Rightarrow z+zx+z{x}^2=1-x+x^2$
$\Rightarrow z{x}^2-x^2+zx+x+z-1=0$
$\Rightarrow x^2(z-1)+x(z+1)+(z-1)=0$
For real x, $B^2-4AC\ge 0$
$\Rightarrow (z+1{)}^2-4(z-1\left)\right(z-1)\ge 0$
$\Rightarrow z^2+2z+1-4z^2+8z-4\ge 0$
$\Rightarrow -3z^2+10z-3\ge 0\Rightarrow -3z^2+9z+z-3\ge 0$
$\Rightarrow -3z(z-3)+1(z-3)\ge 0$
$\Rightarrow (z-3)(-3z+1)\ge 0\Rightarrow \frac{1}{3}\le z\le 3$
$\therefore$ minimum value of $z=\frac{1}{3}$