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

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Aicharya · Accepted Answer

On differentiating w.r.t. $x$, we get $f^{\prime}(x)=\frac{\left(x^2+x+1\right)(2 x-1)-\left(x^2-x+1\right)(2 x+1)}{\left(x^2+x+1\right)^2}$ for maximum or minimum, put $f^{\prime}(x)=0$ $\begin{array}{lc} \Rightarrow & \left(x^2+x+1\right)(2 x-1)-\left(x^2-x+1\right)(2 x+1)=0 \\ \Rightarrow & x^2+x-1-\left(-x^2+x+1\right)=0 \\ \Rightarrow & 2 x^2-2=0 \Rightarrow x= \pm 1 \end{array}$ Now, $f^{\prime}(c)=\frac{2 x^2-2}{\left(x^2+x+1\right)^2}$ Again differentiating, we get $\begin{aligned} & \left(x^2+x+1\right)^2(4 x)-\left(2 x^2-2\right) \\ & f^{\prime \prime}(x)=\frac{2\left(x^2+x+1\right)(2 x+1)}{\left(x^2+x+1\right)^4} \end{aligned}$ at $x=1, f^{\prime \prime}(x)>0$ Therefore it is minimum at $x=1$ Put $x=1$ in equation (i), we get $f(1)=\frac{1-1+1}{1+1+1}=\frac{1}{3}$ $\therefore$ The minimum value is $\frac{1}{3}$.

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

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