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I am trying to prove this function $y = \frac{x} {x+1}$ is increasing. I tried doing by induction: let $n = 1$, then clearly $\frac 1 2 \leq \frac 2 3$ so it's true. But I get stuck on the inductive step, where I'm not even sure how to manipulate the fraction. I have tried multiplying it by various things like $\frac {x+1} {x+1}$ but I guess some more clever manipulation is needed. Any hints would be appreciated here

José Carlos Santos
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    Induction would only prove it is increasing on the natural numbers, you need to show it is true on real numbers.. Hint: Add and subtract one in the numerator. – Alan Sep 09 '21 at 17:50
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    $\frac{1}{1+x}$ is decreasing – The One Sep 09 '21 at 17:51
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    Hint : $y = 1 - \frac{1}{x+1}.$ As $x$ increases, what is happening to the fraction $\frac{1}{x+1}$? – user2661923 Sep 09 '21 at 17:52
  • @user2661923 as $x$ increases, the fraction gets smaller and $y$ gets bigger - I get it now, thank you all! –  Sep 09 '21 at 17:53
  • If you can use calculus, you can show that the derivative is always positive – Vasili Sep 09 '21 at 17:53
  • Can only use induction on natural numbers. The real number equivalence is to prove that $x< y \implies \frac x{x+1} < \frac y{y+1}$. We can do that by manipulating $\frac x{x+1}= \frac{x+1 - 1}{x+1} = \frac{x+1}{x+1} -\frac 1{x+1} = 1-\frac 1{x+1}$. So we just need to show that $x <y \implies 1-\frac 1{x+1} < \frac 1-\frac 1{y+1}$. Can you do that? – fleablood Sep 09 '21 at 18:08
  • To reiterate Jose Carlos Santos and user2661923 comments, though, this function isn't actually increasing. consider $-1.1 < -0.9$ but $f(-1.1) = \frac {-1.1}{-1.1+1}=\frac{-1.1}{-0.1}= 11 > 0$ but $f(-0.9) = \frac {-0.9}{-0.9 + 1} = \frac {-0.9}{0.1} = -9 < 0$. So $f(-1.1) > f(-0.9)$ even then $-1.1 < -0.9$. So it is not increasing.... it is increasing on $(-1, \infty)$ and on $(-\infty, -1)$ individually but it is not increasing on the whole. – fleablood Sep 09 '21 at 18:55

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I'm assuming that $x\in(-1,\infty)$. Then, if $y>x$,$$\frac y{y+1}-\frac x{x+1}=\frac{y-x}{(1+x)(1+y)}>0,$$and therefore $\frac y{y+1}-\frac x{x+1}>0\left(\iff\frac y{y+1}>\frac x{x+1}\right)$.

Note, however, that if you consider that the domain of the expression $\frac x{x+1}$ is $\Bbb R\setminus\{-1\}$, then the statement is false: $-2<0$, but$$\frac{-2}{-2+1}=2>0=\frac0{0+1}.$$

José Carlos Santos
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$\frac{dy}{dx}=\frac{1}{(x+1)^2}>0$ hence the function is increasing

David Quinn
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$y(x)=\frac{x}{x+1} = \frac{x+1-1}{x+1} = \frac{x+1}{x+1} + \frac{-1}{x+1} =1 - \frac{1}{x+1}$

In the last term we have the reciprocal value of $x+1$ which is decreasing.

Multiplying a decreasing expresssion with a negative number like $-1$ makes it increasing.

And adding the constant $1$ does not change the direction, so the complete rhs is overall increasing.

Maksim
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