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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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    What have you tried either this way or your own? What is the most obvious thing you can do for an "algebraic" proof?2012-09-06

2 Answers 2

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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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    Ken, Welcome. You might want to look at [this helpful introduction to $\LaTeX$](http://meta.math.stackexchange.com/questions/5020/tex-latex-mathjax-basic-tutorial-and-quick-reference)2012-09-06
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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 $z$)2012-09-07