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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?

xzhu
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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 06 '12 at 21:08
  • @ThomasAndrews: Thanks for your advice but IMHO that makes it even more difficult to prove. – xzhu Sep 06 '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 06 '12 at 21:15
  • Related : https://math.stackexchange.com/questions/309198/if-dx-y-is-a-metric-then-fracdx-y1-dx-y-is-also-a-metric – Arnaud D. Sep 23 '20 at 17:17

3 Answers3

21

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 $(1 + x)(1 + y)(1 + z)$.

robjohn
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Ken Pledger
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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 $z=x+y$.

Rijul Saini
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Since $f(x)=\frac{x}{x+1}=1-\frac1{x+1}$ is increasing we can consider the case $a,b< c$ then

$$\frac{c}{c+1}< \frac{a+b}{c+1}=\frac{a}{c+1}+\frac{b}{c+1}<\frac{a}{a+1}+\frac{b}{b+1}$$

user
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