Inequality. Further solutions?

Question

For the community of inequalities.

If $z\leq x+y$ ,And $x, y, z \neq -1$, prove :

$$ \frac{z}{1+z} \leq \frac{x}{1+x} + \frac{y}{1+y}. $$

I have this proof, but I feel there should be other simpler (direct) proofs

My attempt:

$$ z(1-z+z^2-\dots)\leq (x+y)(1-z+z^2-\dots), $$

now the idea is that either $(1-z+z^2-\dots)\leq(1-x+x^2-\dots)$ and $(1-z+z^2-\dots)\leq(1-y+y^2-\dots)$ are true, and thus

$$ z(1-z+z^2-\dots)\leq x(1-x+x^2-\dots)+y(1-y+y^2-\dots), $$

which is exactly

$$ \frac{z}{1+z}\leq\frac{x}{1+x} + \frac{y}{1+y} $$.

Note: I have added the above condition for $x,y,z$ to define The LHS and RHS of inequality

Answer 1

Hint: If $-1< a\leq b$ then $${a\over 1+a}\leq {b\over 1+b}$$ and $$ {x\over 1+x+y} \leq {x\over 1+x}$$

Answer 2

The function $f : \mathbb{R} \to \mathbb{R}$ given by $f(t) = \frac{t}{1+t}$ is strictly increasing:

$$f'(t) = \frac{1}{(1+t)^2} > 0$$

Therefore, if we assume $x, y, z \ge 0$ and $z \le x + y$ then

$$\frac{z}{1+z} = f(z) \le f(x + y) = \frac{x+y}{1+x+y} = \frac{x}{1+x+y} + \frac{y}{1+x+y} \le \frac{x}{1+x} + \frac{y}{1+y}$$

Answer 3

$x,y,z \geq 0$ so $0 \leq xy(2+z) $ and $z \leq x+y $, add these \begin{eqnarray*} z &\leq& x+y+2xy+xyz \\ \color{blue}{zx+zy+zxy}+z &\leq& x+y+2xy+xyz +\color{blue}{zx+zy+zxy}\\ z(1+x)(1+y) &\leq& \underbrace{(x+y+2xy)}_{x(1+y)+y(1+x)}(1+z) \\ \frac{z}{1+z} &\leq& \frac{x}{1+x} +\frac{y}{1+y}. \end{eqnarray*}

Answer 4

For positives $x$ and $y$ we obtain: $$\frac{x}{1+x}+\frac{y}{1+y}>\frac{x}{1+x+y}+\frac{y}{1+x+y}=\frac{x+y}{1+x+y}=$$ $$=1-\frac{1}{1+x+y}\geq1-\frac{1}{1+z}=\frac{z}{1+z}.$$