$\frac{c}{1+c}\leq\frac{a}{1+a}+\frac{b}{1+b}$ for $0\leq c\leq a+b$ 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?

$$\frac{a}{1+a}+\frac{b}{1+b}\ge\frac{a}{1+a+b}+\frac{b}{1+b+a}=\frac{a+b}{1+a+b}=\frac{1}{1+\frac{1}{a+b}}\ge\frac{1}{1+\frac1c}=\frac{c}{1+c}$$
