I am sorry if I am asking something too specific and not useful to the general public, but I am stuck at proving that the following function is increasing for all $t\geq 1$ at any given $y$ such that $1\leq y \leq t$:

$F(t) = \frac{t^2-1}{3(2t-1)} - \frac{y}{2t-1}(4t-y-1) +2t(1-(1-\frac{1}{t})^y)$.

I have tried plotting the function, and it seems to be increasing in the specified domain. I have also tried the first derivate, the graph of which is also positive. But how do I prove this formally without the use of graphs?

Thank you in advance for trying and answering.


Promote $t$ to a linear function $t(x)=y+x$ with $y$ constant and domain $x \geq 0$.

This is in agreement with the relations $1 \leq y \leq t$.

Now start with setting $$y=1$$

The domain of $F$ is $[y,\infty)=[1,\infty)$ for now.

$$F(t) = \frac{t^2-1}{3(2t-1)} - \frac{1}{2t-1}(4t-1-1) +2t(1-(1-\frac{1}{t})^1)= \frac{t^2-1}{3(2t-1)} - 2 + 2 = \frac{t^2-1}{3(2t-1)}= \frac{(1+x)^2-1}{3(1+2x)}=x \frac{2+x}{3+6x}$$

$F(y)=F(1)=(F \circ t)(0) = 0$ is easily obtained.

For $x>0$: $$F(t)=F(y+x)=F(1+x)=\frac{2+x}{\frac{3}{x}+6}$$

Clearly $F$ is increasing as $t=y+x=1+x $ is increasing. It is easy to see the other two terms are increasing with $y$ and, and then fixing $y$ again they increase with $t$.

  • $\begingroup$ Thank you, I really appreciate your time to answer this question. However, I am unable to fully understand the last argument in your answer: Now for any larger $y>1$, this is just a fixed increase to $t=y+x$ and therefore $F(t)$ is increasing in $[y,\infty)$ with the original $y≥1$ relation. $\endgroup$ – Kunal Kumar Jan 31 '18 at 9:50

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