$\displaystyle f(x)=\frac {x-5}{x^2-5x+1992}$For which $\displaystyle x$ does it have its maximum?
Although you need to find a solution without using calculus, that doesn’t mean you can’t use calculus to find what the maximum value might be. The problem should be easier if you know what the maximum value is.
Using calculus, I found that the maximum value occurs at $\displaystyle x=5+\sqrt{1992}$. So the problem is to use non-calculus methods to prove that $\displaystyle \frac{x-5}{x^2-5x+1992}\leq\mathrm{f}\left(5+\sqrt{1992}\right)$ for all $\displaystyle x\in\mathbb{R}$.
$\displaystyle \mathrm{f}\left(5+\sqrt{1992}\right)$ should simplify to $\displaystyle \frac{1}{5+2\sqrt{1992}}$. So the inequality becomes
$\displaystyle \color{white}.\quad.$ $\displaystyle \frac{x-5}{x^2-5x+1992}\ \leq\ \frac{1}{5+2\sqrt{1992}}$
$\displaystyle \Leftrightarrow\ (x-5)(5+2\sqrt{1992})\ \leq\ x^2-5x+1992$ (note that $\displaystyle x^2-5x+1992>0$ for all x)
$\displaystyle \Leftrightarrow\ 5x-25+2x\sqrt{1992}-10\sqrt{1992}\ \leq\ x^2-5x+1992$
$\displaystyle \Leftrightarrow\ 0\ \leq\ x^2-2(5+\sqrt{1992})x+25+10\sqrt{1992}+1992\ =\ \left(x-[5+\sqrt{1992}]\right)^2$
QED.
The simple answer was:
$\displaystyle f(x)=\frac {x-5}{x^2-5x+1992}=\frac{1}{x+\frac{1992}{x-5}}=\frac{1}{x-5+\frac{1992}{x-5}+5}\le\frac{1}{4\sqrt{498}+5}$ by AM-GM
Equality occurs when $\displaystyle x-5=\frac{1992}{x-5}\implies x=2\sqrt{498}+5$