Inequality involving maximum of continuous functions
In general, if $d$ is a metric, then so is $\tilde{d}=\frac{d}{1+d}$. This is then just the triangle inequality for the metric $\tilde{d}$ where $d(f,g)=\max_{x\in K}||f(x)-g(x)||$. This is a standard way to define a bounded metric on a given metric space.
To prove the triangle inequality (the hard part of proving $\frac{d}{1+d}$ defines a metric to begin with), examine the function $t\mapsto \frac{t}{1+t}$.
This boils down to
Let $f$ be a suitable function. If $a,b,c\ge 0$ and $c\le a+b$, then $f(c)\le f(a)+f(b)$.
Here "suitable" means
- $ f$ is non-decreasing on $[0,\infty)$
- $f(x+y)\le f(x)+f(y)$
Indeed, from these properties, we obtain $f(c)\le f(a+b)\le f(a)+f(b)$. In your question, $f(x)=\frac x{1+x}=1-\frac1{1+x}$. This is clearly non-decreasing for $x\ge0$. And $$f(x)+f(y)=\frac x{1+x}+\frac y{1+y}=\frac{x+y+2xy}{1+x+y+xy}\ge\frac{x+y+xy}{1+x+y+xy}=f(x+y+xy)\ge f(x+y) $$