For all real $x$, the minimum value of the function $f(x) = \frac{1 - x + x^2}{1 + x + x^2}$ is
Show Hint
Whenever you see a rational function of the symmetric form $\frac{x^2 - kx + 1}{x^2 + kx + 1}$, the minimum and maximum values are always reciprocals of each other! Evaluating the function at quick testing points like $x=1$ gives $f(1) = \frac{1}{3}$, which instantly highlights $\frac{1}{3}$ as a critical value bound.
Step 1: Set the function equal to $y$.
Let $y = \dfrac{1 - x + x^2}{1 + x + x^2}$. We want the least value $y$ can take for real $x$. Step 2: Cross-multiply to form a quadratic in $x$.
$y(1 + x + x^2) = 1 - x + x^2 \Rightarrow (y-1)x^2 + (y+1)x + (y-1) = 0$. Step 3: Demand real $x$.
For real $x$ the discriminant must satisfy $\Delta \ge 0$: $(y+1)^2 - 4(y-1)^2 \ge 0$. Step 4: Expand the discriminant.
$(y^2 + 2y + 1) - 4(y^2 - 2y + 1) = -3y^2 + 10y - 3 \ge 0$. Step 5: Solve the inequality.
Multiply by $-1$: $3y^2 - 10y + 3 \le 0 \Rightarrow (3y - 1)(y - 3) \le 0$, giving $\dfrac{1}{3} \le y \le 3$. Step 6: Read off the minimum.
The smallest allowed value is $y = \dfrac{1}{3}$.
\[ \boxed{\dfrac{1}{3}} \]