What $x$ makes $\frac{x}{(a^2 + x^2)}$ maximum? The problem (Calculus Made Easy, Exercises IX, problem 2 (page 130)) is:

What value of $x$ will make $y$ a maximum in the equation
$$y = \frac{x}{(a^2 + x^2)}$$

I successfully differentiate, equate to zero, and wind up with

$$x^2 = a^2$$

Which gives me the answer of

$$x = a$$

This is correct.  But why isn't $x = -a$ also correct?

Using AM-GM: $a^2+b^2\ge 2ab$ or $a^2+b^2\ge 2|a||b|$.
For $x\ne 0$:
$$\frac{x}{a^2 + x^2}\le \frac{|x|}{a^2+x^2}\le \frac{|x|}{2|a||x|}=\frac1{2|a|},$$
the equality occurs for $x=a>0$ or $x=-a>0$.
If $a=0$, then $y=\frac1x$ does not have a maximum.
