Question

If $x$ real then the minimum value of $\frac{x^2-x+1}{x^2+x+1}$ is __________.

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

Answer 1

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

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

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

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

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

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

$\frac{dy}{dx}=\frac{(x^2+x+1)\times 2-\left(2x\right)(2x+1)}{{(x^2+x+1)}^2}$

$\frac{dy}{dx}=\frac{2(x^2+x+1-x(2x+1))}{{(x^2+x+1)}^2}$

$\frac{dy}{dx}=\frac{2(x^2+x+1-2x^2-x)}{{(x^2+x+1)}^2}$

$\frac{dy}{dx}=\frac{2(-x^2+1)}{{(x^2+x+1)}^2}$

Maximum value occurs when $\frac{dy}{dx}=0$

$\Rightarrow \frac{2(-x^2+1)}{{(x^2+x+1)}^2}=0$

$\Rightarrow -x^2+1=0$

$\Rightarrow x^2-1=0$

$\Rightarrow (x-1)(x+1)=0$

$\therefore x=-1,1$

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

For $x=1,y=\frac{2x}{x^2+x+1}=\frac{2}{1+1+1}=\frac{2}{3}$

$\therefore z_{min}=1-y_{max}=1+2=3$ for $y=-2$

and $z_{min}=1-y_{max}=1-\frac{2}{3}=\frac{1}{3}$ for $y=\frac{2}{3}$

Since $\frac{1}{3}<3$

Thus, the least value is $\frac{1}{3}$