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With $x+y\ge z$ $(x,y,z\ge0)$, prove that: $$\frac{x}{1+x}+\frac{y}{1+y}\ge\frac{z}{1+z}$$

I'm aware that using analytic view this is easy since $f(x)=\frac{x}{1+x}$ is concave in $[0,\infty)$. However I want to prove it using merely algebraic techniques. Is that possible?

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Write $\frac{w}{1+w}=1-\frac{1}{1+w}$ for $w=x,y,z$ and then you need to show $\frac{1}{1+x}+\frac{1}{1+y}\leq 1 + \frac{1}{z+1}$. Proving this is fairly direct algebraically. –  Thomas Andrews Sep 6 '12 at 21:08
    
@ThomasAndrews: Thanks for your advice but IMHO that makes it even more difficult to prove. –  Voldemort Sep 6 '12 at 21:11
    
What have you tried either this way or your own? What is the most obvious thing you can do for an "algebraic" proof? –  Thomas Andrews Sep 6 '12 at 21:15
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2 Answers

up vote 18 down vote accepted

The conditions certainly give $xyz + 2xy + x + y \ge z$

which is a simplification of $x(1 + y)(1 + z) + (1 + x)y(1 + z) \ge (1 + x)(1 + y)z$.

Now just divide all that by $(1 + x)(1 + y)(1 + z)$.

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Why the downvotes? –  Cocopuffs Sep 6 '12 at 21:22
    
Probably 'cause he didn't use $\LaTeX$. –  Voldemort Sep 6 '12 at 21:23
    
This looks fine to me. So I ask, too: why the downvotes? –  robjohn Sep 6 '12 at 21:23
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@Voldemort: I certainly hope not. That is no way to greet a newcomer. Instead of downvoting, or at least in addition to downvoting, suggest the use of $\LaTeX$. –  robjohn Sep 6 '12 at 21:25
    
@robjohn: I agree. That's why I gave him upvote and a suggestion. –  Voldemort Sep 6 '12 at 21:27
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How about this? As Thomas Andrews observed, write $\frac{w}{1+w}=1−\frac{1}{1+w}$ for $w=x,y,z$ and then you need to show $\frac{1}{1+x} + \frac{1}{1+y} \le 1 + \frac{1}{z+1}$.

Now, $$1 + \frac{1}{z+1} \ge 1 + \frac{1}{x+y+1} = \frac{x+y+2}{x+y+1} \ge \frac{(x+1)+(y+1)}{1+x+y+xy} = \frac{1}{1+x} + \frac{1}{1+y}$$ and we're done. Equality holds when one of $x,y$ is $0$, and $z=x+y$.

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(so equality holds when one of $x$ or $y$ is zero and the other is equal to $z$) –  Ben Millwood Sep 7 '12 at 1:00
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