Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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

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, $0 < f \le g+h \implies$ $$ \frac{f}{f+1} = p\Bigl(f\left(x\right)\Bigr) \le p\Bigl(g\left(x\right)+h\left(x\right)\Bigr) = \frac{g+h}{g+h+1} = \frac{g}{g+h+1} + \frac{h}{g+h+1} < \frac{g}{g+1} + \frac{h}{h+1} \,. $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.