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If I have functions $f,g,h > 0$ and $f\le g+h$ then:

$$\frac f{f+1}\le \frac g{g+1}+\frac h{h+1}, x \in R$$

I have been trying to find out whether it's true or not but I haven't succeeded.

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Assuming that you mean real-valued functions have you checked it for constant functions, (i.e. $f,g,h$ are real numbers)? –  Nils Matthes Jun 6 '12 at 10:39
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have you tried f=g+h, because x/(1+x) is increasing provided x>0. –  Yimin Jun 6 '12 at 10:52
    
The real number case Nils mentions is really the whole problem. –  Jonas Meyer Jun 6 '12 at 10:54
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1 Answer 1

As Yimin says, the function $p(x)=\frac{x}{x+1}=1-\frac1{x+1}$ is increasing for $x > 0$ (it's a flip and shift of the double hyperbola $y=\frac1x$, with resulting asymptotes $x=-1$ and $y=1$). Therefore, $0 < f \le g+h \implies$ $$ \frac{f}{f+1} = p\Bigl(f\left(x\right)\Bigr) \le p\Bigl(g\left(x\right)+h\left(x\right)\Bigr) = \frac{g+h}{g+h+1} = \frac{g}{g+h+1} + \frac{h}{g+h+1} < \frac{g}{g+1} + \frac{h}{h+1} \,. $$

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