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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?

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marked as duplicate by Martin R, Michael Rozenberg calculus Oct 30 '18 at 8:38

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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 $x,y,z\geq 0$.

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    $\begingroup$ 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 $$x,\;\;y,\;\;z,\;\;x+y-z$$ $\endgroup$ – quasi Oct 29 '18 at 23:46
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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.

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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 $b=0$.

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