If $x\leq y + z$, then how can I show that $x/(1+x) \leq 1/(1+y) +1/(1+z)$? I am trying to do this algebraically by dividing through the initial expression by $1+x$. But, I am not get anywhere with it.

  • $\begingroup$ Are you dealing with non-negative numbers? $\endgroup$ – mickep Oct 15 '15 at 20:10
  • 2
    $\begingroup$ Is there a typo? It's false in general, e.g. take $x=1$ and $y=z=10$. Probably you want $y$ and $z$ in the numerators on the RHS. $\endgroup$ – Bungo Oct 15 '15 at 20:18

I assume you mean $$\frac{x}{1+x} \leq \frac{y}{1+y} + \frac{z}{1+z}$$ First note that $\displaystyle x \mapsto \frac{x}{1+x}$ is an increasing function on $[0,\infty)$. To see this, assume that $0 \leq x \leq y$. Then add $xy$ to both sides of $x \leq y$ to obtain $$x(1+y) \leq y(1+x)$$ Divide both sides by $(1+x)(1+y)$ to conclude that $$\frac{x}{1+x} \leq \frac{y}{1+y}$$ Now suppose that $0 \leq x \leq y + z$. By above inequality, it follows that $$\frac{x}{1+x} \leq \frac{y+z}{1 + y + z}$$ Manipulate the RHS as follows: $$\begin{aligned} \frac{y+z}{1 + y + z} &= \frac{y}{1 + y + z} + \frac{z}{1 + y + z} \\ &\leq \frac{y}{1+y} + \frac{z}{1+z} \\ \end{aligned}$$


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