Strictly increasing function If $f(x)$ is a continuous function on $\mathbb R$, and $|f(-x)|< |f(x)|$ for all $x>0$. Does it imply that $|f(x)|$ is strictly increasing on $(0,\infty)$?
I tried to use the definition: let $a,b \in (0,\infty)$ with $a<b$, we need to show that $|f(a)|<|f(b)|$. We have 
$|f(-a)|< |f(a)|$ and $|f(-b)|< |f(b)|$, and I don't know how to proceed!

While this question is long answered, all answers use some piecewise function. Therefore I want to add one additional counterexample which isn't piecewise:
$$f(x) = \frac{\mathrm e^x}{1+\mathrm e^{x^2}}$$
Clearly for $x\in(0,\infty)$ we have $\lvert f(-x)\rvert < \lvert f(x)\rvert$: Since the function is positive, the absolute value bars have no effect, and since $x^2=(-x)^2$, the condition reduces to $\mathrm e^{-x}<\mathrm e^x$ which is true for $x>0$ because the exponential function is strictly increasing. There's also no question that the function is continuous. On the other hand, $f(0)=1/2$, but $\lim_{x\to\infty}f(x)=0$, so it's also obvious that it is not strictly increasing.
Here's a graph of the function:

