Question

If $x$ is real, then the minimum value of $y=\frac{x^2-x+1}{x^2+x+1}$ is

• A. $3$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $2$

Answer 1

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

$y=\frac{x^2-x+1}{x^2+x+1}$

$y{x}^2+yx+y=x^2-x+1$

$\Rightarrow x^2(y-1)+x(y+1)+(y-1)=0$

$x$ is real, so $D\ge 0$,

$⟹(y+1{)}^2-4(y-1{)}^2\ge 0$

$⟹y^2+1+2y-4[y^2+1-2y]\ge 0$

$⟹y^2+1+2y-4y^2-4+8y\ge 0$

$⟹-3y^2+10y-3\ge 0$

$⟹3y^2-10y+3\le 0$

$⟹3y^2-9y-y+3\le 0$

$⟹3y(y-3)-(y-3)\le 0$

$⟹(3y-1)(y-3)\le 0$

$⟹y\le 3$ and $y\ge \frac{1}{3}$

$⟹\frac{1}{3}\le y\le 3$

Hence, $\frac{1}{3}$ is the minimum value of $y$.