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I am trying to prove the following:

$$\frac{a}{1+a} < \frac{b}{1+b} + \frac{c}{1+c}$$

given that $a, b, c > 0$ and $a < b+c$. I tried various rearrangements but can't seem to get anywhere with it.

Michael Rozenberg
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user115697
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    This is a proof of the triangle inequality for the metric $\rho(x,y)=d(x,y)/(1+d(x,y))$ given the metric $d(x,y)$. – abnry Feb 11 '14 at 23:09

4 Answers4

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Function $f(x)=\frac{x}{1+x}$ is increasing (it is easy to check first derivative). Therefore $f(a)<f(b+c)$, for $a<b+c$. Because of that, we have $$\frac{a}{1+a}<\frac{b+c}{1+b+c}=\frac{b}{1+b+c}+\frac{c}{1+b+c},$$ from where we get (since $a,b,c>0$): $$\frac{a}{1+a}<\frac{b}{1+b}+\frac{c}{1+c}.$$

alans
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By C-S $$\frac{b}{1+b}+\frac{c}{c+1}=\frac{b^2}{b^2+b}+\frac{c^2}{c^2+c}\geq\frac{(b+c)^2}{b^2+c^2+b+c}\geq\frac{(b+c)^2}{(b+c)^2+b+c}=$$ $$=\frac{b+c}{b+c+1}=1-\frac{1}{b+c+1}>1-\frac{1}{a+1}=\frac{a}{a+1}.$$ Done!

Michael Rozenberg
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Let us prove the equivalent inequality

$$\frac{b}{1+b} + \frac{c}{1+c} - \frac{a}{1+a} > 0.$$

Since $\frac{n}{1+n} = 1 - \frac{1}{1+n}$, observe that

$$\begin{align*} \frac{b}{1+b} + \frac{c}{1+c} - \frac{a}{1+a} &= 1 - \frac{1}{1+b} - \frac{1}{1+c} + \frac{1}{1+a}\\ &> 1 - \frac{1}{1+b} - \frac{1}{1+c} + \frac{1}{1+b + c + bc}\\ &= \left ( 1 - \frac{1}{1+b} \right ) \left ( 1 - \frac{1}{1+c} \right ) \\ &> 0. \end{align*}$$

The first inequality used is $$\frac{1}{1+a} > \frac{1}{1 + b + c + bc},$$ which is true since $a < b + c$ and $0 < bc$.

The final inequality holds since $b, c > 0$, so both terms in parentheses are positive.

Ben
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$$\frac{b}{1+b} + \frac{c}{1+c}>\frac{a}{a+1}$$

$$\frac{b+c+2bc}{1+b+c+bc}>\frac{a}{a+1}$$ Since $a,b,c>0$, $$ab+ac+2abc+b+c+2bc>a+ab+ac+abc$$

$$abc+2bc+b+c>a$$

Bobby
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