Calculating the infimum of $f(t)=\frac{\epsilon t+(1-\frac1{m})\frac{t}{t-1}}{1-\frac1{m}+\frac{t}{m}}$ for $t>1$ Suppose $\epsilon$ is a positive number and $m$ is an positive integer. Define a function 
$$f(t) = \frac{\epsilon t + (1-\frac{1}{m})\frac{t}{t-1}}{1-\frac{1}{m}+\frac{t}{m}}$$
How to get the infnimum  of $f(t)$ when $t> 1$?
 A: One can write $f(t)=\frac{N(t)}{D(t)}$ with
$$
N(t)=m\epsilon(t^2-t)+(m-1)t, \ D(t)=(t-1)(t+m-1) \tag{1}
$$
Thus, $f'(t)=\frac{N'(t)D(t)-N(t)D'(t)}{D(t)^2}$. When we expand the numerator, we obtain $N'(t)D(t)-N(t)D'(t)=(m-1)M(t)$ with
$$
M(t)= (m\epsilon-1)t^2-2m\epsilon t +(1-m+m\epsilon)  \tag{2}
$$
This is a quadratic polynomial (unless $m\epsilon-1=0$), with discriminant
$$
\mu = (2m\epsilon)^2-4(m\epsilon-1)1-m+m\epsilon)=
4 (m^2 \epsilon + 1-m) =4m^2 \Bigg(\epsilon-\bigg(\frac{1}{m}-\frac{1}{m^2}\bigg)\Bigg) \tag{3}
$$
We can now distinguish several cases.
Case 1 : $\epsilon \leq \frac{1}{m}-\frac{1}{m^2}$
Case 2 : $\frac{1}{m}-\frac{1}{m^2} \lt \epsilon \lt \frac{1}{m}$
Case 3 : $\epsilon=\frac{1}{m}$
Case 4 : $\frac{1}{m} \lt \epsilon$
In case 1, $M$ has a negative discriminant and a negative leading term, so $M$ is always negative. So $f$ is decreasing and its infimum is its limit at $+\infty$, namely $m\epsilon$.
In case 2, $M$ has a positive discriminant and a negative leading term. It has two roots, $\alpha=\frac{-m\epsilon}{1-m\epsilon}-\frac{\sqrt{\mu}}{2(1-m\epsilon)}$ and $\beta=\frac{-m\epsilon}{1-m\epsilon}+\frac{\sqrt{\mu}}{2(1-m\epsilon)}$. We see that $\alpha \leq 0$ ; on the other hand, from (3) we deduce that $\mu \lt 4$, so $\sqrt{\mu} \lt 2$ and hence $\beta \lt 1$. So both the roots are outside $(1,\infty)$, $f$ is decreasing again and its infimum is $m\epsilon$ as in case 1.
In case 3, we have $M(t)=2-m-2t \leq -m \lt 0$ so the infimum is still $m\epsilon$.
In case 4, $M$ has a positive discriminant and a positive leading term. It has two roots, $\gamma=\frac{m\epsilon}{m\epsilon-1}-\frac{\sqrt{\mu}}{2(m\epsilon-1)}$ and $\delta=\frac{m\epsilon}{m\epsilon-1}+\frac{\sqrt{\mu}}{2(m\epsilon-1)}$. We see that $\delta \geq \frac{m\epsilon}{m\epsilon-1} \gt 1$ ; on the other hand, from (3) we deduce that $\mu \gt 4$, so $\sqrt{\mu} \gt 2$ and hence $\gamma \lt 1$. So there is exactly one root in $(1,\infty)$ namely $\delta$, $f$ is decreasing on $(1,\delta)$ and increasing on $(\delta,\infty)$, and therefore reaches its infimum at $\delta$.
