When proving that if $d$ is a metric, then $d'(x,y)=\dfrac{d(x,y)}{1+d(x,y)}$ is also a metric, I have to prove the inequality:
$$\dfrac{c}{1+c}\leq\frac{a}{1+a}+\frac{b}{1+b}$$ for $0\leq c\leq a+b$. This is obvious by expanding, but is there a nicer way to see why it is true?
$$\frac{a}{1+a}+\frac{b}{1+b}\ge\frac{a}{1+a+b}+\frac{b}{1+b+a}=\frac{a+b}{1+a+b}=\frac{1}{1+\frac{1}{a+b}}\ge\frac{1}{1+\frac1c}=\frac{c}{1+c}$$
$\displaystyle\frac{d(x,z)}{1+d(x,z)}-\frac{d(x,y)}{1+d(x,y)}=\frac{(d(x,z)-d(x,y))}{(1+(d(x,z))(1+d(x,y))}$
We havefrom $\Delta $ inequality,
$(1+(d(x,z))(1+d(x,y))\ge1+d(x,z)+d(x,y)\ge 1+d(y,z)\tag 1$
And we have from $\Delta $ inequality $d(x,z)-d(x,y)\le d(y,z)\tag 2$
And as $1+d(y,z)\ne 0$ we have $\displaystyle\frac{1}{1+d(y,z)}\ge \frac{1}{(1+d(x,z))(1+d(x,y))}$ from $(1)$
Multiplying $(2) $ and $(1)$ we have
$$\frac{d(x,z)}{1+d(x,z)}-\frac{d(x,y)}{1+d(x,y)}=\frac{(d(x,z)-d(x,y))}{(1+(d(x,z))(1+d(x,y))}\le \frac{d(y,z)}{1+d(y,z)}$$
$$\Rightarrow \frac{d(x,z)}{1+d(x,z)}-\frac{d(x,y)}{1+d(x,y)}\le \frac{d(y,z)}{1+d(y,z)}$$
$$\Rightarrow \frac{d(x,z)}{1+d(x,z)}\le \frac{d(x,y)}{1+d(x,y)}+ \frac{d(y,z)}{1+d(y,z)}$$
