AM–GM / Bounding Inequality Technique

Fast Inequality Solution (under 45 s)

1. Apply the AM–GM inequality to the denominator:

   $$ x^{2} + 1 ;\ge; 2\sqrt{x^{2}!\cdot!1}=2|x|. $$

2. For $x>0$ the bound becomes $x^{2}+1\ge 2x$. Hence

   $$ f(x) =\frac{12x}{x^{2}+1} ;\le; \frac{12x}{2x} =6. $$

   Equality occurs precisely when $x^{2}+1 = 2x$ ↔ $x=1$.

3. For $x\le 0$ the numerator $12x$ is non‑positive, so $f(x)\le6$.

Therefore the largest possible value of $f(x)$ is achieved at $x=1$.