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

(A) $0$

(B) $1$

(C) $3$

(D) $(1/3)$

Option D is correct

Let $y = \cfrac{{{x^2} - x + 1}}{{{x^2} + x + 1}} \Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{{2(x - 1)(x + 1)}}{{{{({x^2} + x + 1)}^2}}}$

For critical points, we have $\cfrac{{dy}}{{dx}} = 0 \Rightarrow x = 1, - 1$

At $x = 1,\cfrac{{dy}}{{dx}}$ changes its sign from $- ve$ to $+ ve$.

At $x = - 1,\cfrac{{dy}}{{dx}}$ changes its sign from $+ ve$ to $- ve$.

Thus, $y$ is minimum at $x = 1$.

Therefore minimum value of $y = \cfrac{{1 - x + {x^2}}}{{1 + x + {x^2}}}$ at $x = 1$ is $\cfrac{{1 - 1 + 1}}{{1 + 1 + 1}} = \cfrac{1}{3}$.