Can I prove this inequality algebraically?

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?

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

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 $z=x+y$.