Minimum Value of Quotient Function

For $x \neq 1, -1$: $\frac{f(x)}{g(x)} = \frac{(x^2+1)/(x^2-1)}{(x+1)/(x-1)} = \frac{x^2+1}{(x+1)^2}$. Let $y = \frac{x^2+1}{(x+1)^2} \implies y(x^2+2x+1) = x^2+1 \implies x^2(y-1) + 2yx + (y-1) = 0$. Discriminant $D \ge 0 \implies 4y^2 - 4(y-1)^2 \ge 0 \implies 4y^2 - 4(y^2 - 2y + 1) \ge 0 \implies 8y - 4 \ge 0 \implies y \ge \frac{1}{2}$. Check special points: For $x = 1$: $f(1) = 1$, $g(1) = 3 \implies \frac{f(1)}{g(1)} = \frac{1}{3}$. For $x = -1$: $f(-1) = 1$, $g(-1) = 0$, undefined. Since $\frac{1}{3} < \frac{1}{2}$, the minimum possible value is $\frac{1}{3}$.