Hello! I have this problem.
$\displaystyle f(x) = \frac {x^2}{x^2+1} >= \frac {1}{2} $
Here is a slightly different approach:
Since $\displaystyle x^2 + 1 \geq 1$ the inequality becomes:
$\displaystyle x^2\geq\frac12 x^2 + \frac12$
$\displaystyle \frac12 x^2- \frac12\geq 0~\implies~x^2-1\geq 0~\implies~(x+1)(x-1)\geq 0$
A product of 2 factors is positive if both factors have the same sign:
$\displaystyle x+1\geq 0\wedge x-1\geq 0~\vee ~x+1\leq 0\wedge x-1\leq 0$
$\displaystyle x\geq -1\wedge x\geq 1~\vee ~x\leq -1\wedge x\leq 1$
$\displaystyle x\geq 1~\vee ~x\leq -1$
Rewrite it as $\displaystyle \frac{x^2}{x^2+1}- \frac{1}{2}= \frac{2x^2}{2(x^2+1}- \frac{x^2+ 1}{2(x^2+ 1}= \frac{x^2- 1}{x^2+ 1}\ge 0$
A fraction is positive if and only if the numerator and denominator have the same sign. This, denominator, $\displaystyle x^2+ 1$, is always positive.