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

• You can also use the monotone increasing property of $\frac{x}{1+x}$ for proving $\frac{d(x,y)}{1+d(x,y)}$ is a metric. – mtm Jun 17 '13 at 17:50

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

• this comment was dumb, sorry, really need to get some sleep. cheers – mm-aops Jun 15 '13 at 13:23
• Very clear, thanks Maisam! – Paul S. Jun 15 '13 at 13:27
• @mm-aops:no problem care free – M.H Jun 15 '13 at 13:28
• @Paul S.:your welcome dear paul – M.H Jun 15 '13 at 13:29

$\displaystyle\frac{d(x,z)}{1+d(x,z)}-\frac{d(x,y)}{1+d(x,y)}=\frac{(d(x,z)-d(x,y))}{(1+(d(x,z))(1+d(x,y))}$

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

$$\frac{d(x,z)}{1+d(x,z)}-\frac{d(x,y)}{1+d(x,y)}=\frac{(d(x,z)-d(x,y))}{(1+(d(x,z))(1+d(x,y))}\le \frac{d(y,z)}{1+d(y,z)}$$

$$\Rightarrow \frac{d(x,z)}{1+d(x,z)}-\frac{d(x,y)}{1+d(x,y)}\le \frac{d(y,z)}{1+d(y,z)}$$

$$\Rightarrow \frac{d(x,z)}{1+d(x,z)}\le \frac{d(x,y)}{1+d(x,y)}+ \frac{d(y,z)}{1+d(y,z)}$$