How can I demonstrate a mathematical proof that the following function $f(t)$
$$f(t)=-\frac{1}{t}\ln\left( e^{\alpha t}\left(2\beta-\frac{3\beta }{2} e^{\frac{\alpha t}{2}}\right)\right)$$
is decreasing in $t$, where $t\geq0$, $0\leq \beta \leq 1$ and $\alpha <0$. since
$f'(t)=\frac{\ln \left(e^{\alpha t} \left(2 \beta -\frac{3}{2} \beta e^{\frac{\alpha t}{2}}\right)\right)}{t^2}-\frac{e^{\alpha (-t)} \left(\alpha e^{\alpha t} \left(2 \beta -\frac{3}{2} \beta e^{\frac{\alpha t}{2}}\right)-\frac{3}{4} \alpha \beta e^{\frac{3 \alpha t}{2}}\right)}{t \left(2 \beta -\frac{3}{2} \beta e^{\frac{\alpha t}{2}}\right)}$
it is always negative.
Is there an importance to alpha and beta values in the proof of decreasing function for t?
Is this proof convincing of the particular situation?
In particular, $$ f(t)=-\frac{1}{t}\ln(\beta e^{\alpha t})=\alpha-\frac{\ln \beta}{t}$$ where $t\geq0$, $0\leq \beta \leq 1$ and $\alpha <0$.
$$f'(t)=\frac{\ln \beta}{t^2}$$ as $0\leq \beta \leq 1$ is smaller than 1 and the logarithm is negative.
Hence $f'(t)$ is negative for any $t$
This is not a complete solution. I did my best, but the problem is difficult because of the mix of algebraic and logarithm. I tried to prove that the derivative is negative, but there is a detail in the end that I could not fix.
$$f'(t)=\frac{2 \log \left(b \left(2 e^{a t}-\frac{3}{2} e^{\frac{3 a t}{2}}\right)\right)+a t \left(\frac{4}{4-3 e^{\frac{a t}{2}}}-3\right)}{2 t^2}$$ As denominator is positive for any $t$ we will discuss the numerator:
First part is negative because when the argument of logarithm is smaller than $1$ for any $t<0$
$b \left(2 e^{a t}-\frac{3}{2} e^{\frac{3 a t}{2}}\right)<1$
Indeed, $a<0$ means that both $e^{at}$ and $e^{\frac{3 a t}{2}}$ are less than $1$ therefore, for any $t>0$ we have $2 e^{a t}-\frac{3}{2} e^{\frac{3 a t}{2}}<\frac{1}{2}$
as $0\le b\le 1$ it follows that the product is smaller than $1$ and the logarithm is negative.
The second part has a factor $at$ which is negative for the conditions on variables and parameters
$$\frac{4}{4-3 e^{\frac{a t}{2}}}-3=-\frac{9 e^{\frac{a t}{2}}-8}{3 e^{\frac{a t}{2}}-4}>0$$
substitute $u=e^{\frac{a t}{2}}$ and solve
$-\dfrac{9 u-8}{3 u-4}>0\to \dfrac{8}{9}<u<\dfrac{4}{3}$
which is $\dfrac{8}{9}<e^{\frac{a t}{2}}<\dfrac{4}{3}$
that is $\log\dfrac{8}{9}<\dfrac{a t}{2}<\log\dfrac{4}{3}$
which for $a<0$ has the solution $0<t<\frac{2 \log \left(\frac{8}{9}\right)}{a}$
That is about $0<t<-\dfrac{0.24}{a}$
This means that in this interval we are sure that the derivative is negative, for greater values of $t$ the second part of the numerator of the derivative is positive and I don't know if it is, in absolute value, greater or smaller than the first part that is negative for any $t$
Hope this helps anyway