# Prove that : $\frac{|a+b|}{1+|a+b|} \leq \frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}$ [duplicate]

Exercise :

Prove that : $$\frac{|a+b|}{1+|a+b|} \leq \frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}$$ for $$a,b \in \mathbb R$$.

Methods I have tried so far include: Using the triangle inequality on the numerator on the left side, but I got an expression which was sometimes too big, so it's impossible.

Using a similar method on the right side, but I got a pretty nasty expression so I don't think that's the way.

Going case by case for every pair of a,b but this is also very long.

Can someone give a hint?

## marked as duplicate by Martin R, Michael Rozenberg calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 30 '18 at 8:38

• – Martin R Oct 30 '18 at 8:00

The pedestrian approach works: with $$x=|a|$$, $$y=|b|$$, and $$z=|a+b|$$, we have $$(x(1+y)+y(1+x))(1+z)-z(1+x)(1+y)=[x+y-z]+2[xy+xyz]\geq0.$$ The last inequality above holds because $$x+y-z\geq 0$$ (triangle inequality) and because $$x,y,z\geq 0$$.

• That's the best approach (+1). The idea is that since there's no information about $a,b$, the only relation between $$x=|a|,\;\;y=|b|,\;\;z=|a+b|$$ is the triangle inequality $x+y\ge z$, so if the claimed inequality is true, the intuition is that it must be a simple consequence of the nonnegativity of the quantities $$x,\;\;y,\;\;z,\;\;x+y-z$$ – quasi Oct 29 '18 at 23:46

Let $$s=|a|$$, and let $$t=|b|$$.

Consider two cases . . .

Case $$(1)$$:$$\;ab\ge 0$$.

Since $$ab\ge 0$$, it follows that $$a,b$$ do not have opposite signs, hence $$|a+b|=s+t$$, so

\begin{align*} &\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}-\frac{|a+b|}{1+|a+b|}\\[4pt] =\;&\frac{s}{1+s}+\frac{t}{1+t}-\frac{s+t}{1+(s+t)}\\[4pt] =\;&\frac{st(2+s+t)}{(1+s)(1+t)(1+s+t)}\\[4pt] \ge\;&\;0\\[4pt] \end{align*} Thus, the claimed inequality holds for case $$(1)$$.

Case $$(2)$$:$$\;ab < 0$$.

Without loss of generality, assume $$s\ge t$$.

Since $$ab < 0$$, it follows that $$a,b$$ have opposite signs, hence $$|a+b|=s-t$$, so \begin{align*} &\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}-\frac{|a+b|}{1+|a+b|}\\[4pt] =\;&\frac{s}{1+s}+\frac{t}{1+t}-\frac{s-t}{1+(s-t)}\\[4pt] =\;&\frac{t\bigl(s(s-t)+2s+2\bigr)}{(1+s)(1+t)(1+(s-t))}\\[4pt] \ge\;&\;0\\[4pt] \end{align*} Thus, the claimed inequality also holds for case $$(2)$$.

Therefore the claimed inequality always holds.

The required inequality is equivalent to $$1+\frac{1}{1+|a+b|}\geq \frac{1}{1+|a|}+\frac{1}{1+|b|}\,.$$ By the Triangle Inequality, $$|a+b|\leq |a|+|b|$$, so we have $$1+\frac{1}{1+|a+b|}\geq 1+\frac{1}{1+|a|+|b|}\,.$$ Therefore, it suffices to show that $$1+\frac{1}{1+|a|+|b|}\geq \frac{1}{1+|a|}+\frac{1}{1+|b|}\,.$$ This is clearly true, as \begin{align}1+\frac{1}{1+|a|+|b|}&=\frac{2+|a|+|b|}{1+|a|+|b|}\geq \frac{2+|a|+|b|}{1+|a|+|b|+|ab|}\\&=\frac{2+|a|+|b|}{\big(1+|a|\big)\big(1+|b|\big)}=\frac{1}{1+|a|}+\frac{1}{1+|b|}\,.\end{align} Hence, $$\frac{|a+b|}{1+|a+b|}\leq \frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}\,,$$ as desired. The equality holds if and only if $$a=0$$ or $$b=0$$.