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how can I prove the following inequality?

$$\frac{\mid x+y\mid}{1+\mid x+y\mid}\leq\frac{\mid x \mid}{1+\mid x\mid}+\frac{\mid y \mid}{1+\mid y \mid}$$

I was trying to prove it by $$0\leq\alpha\leq\beta $$ $$ \frac{\alpha}{1+\alpha}\leq\frac{\beta}{1+\beta}$$ but how do i prove $$ \frac{\alpha}{1+\alpha}\leq\frac{\beta}{1+\beta}$$ in the first place?

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Just multiply throughout by the denominators and cancel terms. –  Calvin Lin Apr 18 at 23:56
    
$f(t) = t/(1+t) = 1 - 1/(1+t)$ is an increasing function of $t \ge 0$. –  Robert Israel Apr 19 at 0:01

3 Answers 3

up vote 2 down vote accepted

$f(x) = \dfrac{x}{1 + x}$ has $f'(x) = \dfrac{1}{(1 + x)^2} > 0$. So $f$ increases on $[0, +\infty)$. Hence:

$\dfrac{|x+y|}{1 + |x+y|} \leq \dfrac{|x| + |y|}{1 + |x| + |y|} = \dfrac{|x|}{1 + |x| + |y|} + \dfrac{|y|}{1 + |x| + |y|} \leq \dfrac{|x|}{1 + |x|} + \dfrac{|y|}{1 + |y|}$

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Fractions algebra:

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

(note that all is positive here!).

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DON!! como estas? :p –  Pedro Arbizu Apr 19 at 0:02
1  
Todo bien por acá, Lemur (el "otro" Don...) :) –  DonAntonio Apr 19 at 0:03

Let $f(t) = \frac{t}{t+1} $ Notice $f' = \frac{1}{(1+t)^2} > 0 $ Hence $f$ is increasing. In particular,

$$ |x+y| \leq|x| + |y| \implies f(|x+y|) \leq f(|x|+|y|) $$

Hence

$$ \frac{ \ |x+y| }{1 + |x+y|} \leq \frac{|y| + |x| }{1 + |x| +|y|} = \frac{|y|}{1 + |x| +|y|} + \frac{|x|}{1 + |x| +|y|} \leq \frac{|y|}{1+|y|} +\frac{|x|}{1+|x|}$$

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