# Triangle inequality on t/(1+t) [closed]

Let $$a,b,c \in \mathbb{R}$$ such that $$a \leq b+c$$

Show that $$\frac{a}{1+a}\leq\frac{b}{1+b}+\frac{c}{1+c}$$

I'm not sure how to prove this, I'd appreciate some help!

• It's not true if negative numbers are allowed. $a= 1$ $b=-\frac 12$ and $c= 2$ fails. If you assume all are positive then... just do the algebra. – fleablood Oct 25 '19 at 17:31

The inequality works only for $$a, b, c \ge 0$$. Since $$f(x) = \frac{x}{1+x}$$ is a monotone increasing function on $$[0, \infty)$$, we have \begin{align*} \frac{a}{1+a} &\le \frac{b+c}{1+b+c} \\ &=\frac{b}{1+b+c} + \frac{c}{1+b+c} \\ &\le \frac{b}{1+b} + \frac{c}{1+c} \end{align*}

It is $$\frac{b}{1+b}+\frac{c}{1+c}-\frac{a}{1+a}={\frac {abc+2\,bc-a+b+c}{ \left( 1+b \right) \left( 1+c \right) \left( 1+a \right) }}>0$$ True, since $$b+c>a$$ and $$a,b,c$$ the sidelength of an triangle.

$$a,, b, c \ge 0$$, then $$\frac{b}{1+b}>\frac{b}{1+b+c} ~~~(1)$$ $$\frac{c}{1+c} \ge \frac{c}{1+b+c} ~~~~(2)$$ adding the we get $$\frac{b}{1+b}+\frac{c}{1+c} \ge \frac{b+c}{1+b+c}~~~(3)$$ Next, $$b+c\ge a \implies \frac{1}{b+c} \le \frac{1}{a} \implies 1+\frac{1}{b+c} \le 1+ \frac{1}{a} \implies \frac{1+b+c}{b+c} \le \frac{1+a}{a}$$ $$\implies \frac{b+c}{1+b+c} \ge \frac{a}{1+a}~~~~(4)$$ By (3) and (4), we prove that $$\frac{b}{1+b}+\frac{c}{1+c} \ge \frac{a}{1+a}.$$

Muck around till you get it.

Assuming all are positive then

$$\frac a{1+a} \le \frac b{1+b} + \frac c{1+c}\iff$$

$$a(1+b)(1+c) \le b(1+c)(1+a) + c(1+b)(1+a)\iff$$

$$a+ ab+ac + abc \le b+ab+bc+abc + c + ac + bc +abc\iff$$

$$a \le b+c +2bc + abc$$ which is true if all are non-negative.

But this isn't necessarily true if some are negative. Example if $$a=1; b=-\frac 12;c= 2$$

Then $$a = 1 > -\frac 12+2-2-1 = b+c + 2b +abc$$

And $$\frac a{1+a} = \frac 12$$ and $$\frac b{1+b} + \frac c{1+c} = -1+ \frac 23=-\frac 13 < \frac 12 =\frac a{1+a}$$.