Consider this inequality with integers $a >0, b> 0, c>0$:
$$126 > 2^{a+1}(2^b(2^c-1) - 3) > 0$$
I am looking for an integer upper limit $K$ for the sum of $a+b+c$ such that:
$$a + b + c < K$$
It is straight forward to find an upper limit for any one value such as $c$. I am not clear how to find $K$ for $a+b+c$. I would suspect that this is doable since it is possible for the case $126 > 2^{a+b+c+1}$
Here's how I would estimate $K$ for $c$ alone:
(1) $2^b(2^c-1) < \dfrac{126}{2^{a+1}} + 3$
(2) $2^c < \dfrac{126}{2^{a+1+b}} + \dfrac{3}{2^{b}} + 1 < \dfrac{126}{2^{a+1+b}} + 3 < 19$
(3) $c \le 4$
How can $K$ be estimated for two such terms? How can $K$ be estimated for all $3$?
Edit: Updated the bound based on feedback from @player3236
One way to approach these types of problems is consider the relative rates of change of the function for each variable, i.e., compare each of the partial derivatives. Note this only works, by itself, if there is always the same relative ordering of derivatives for all of the variable values in the domain, at least to start with to have a smallest derivative value.
If there's an always smallest derivative, note if it's always $\le 0$ for all values of that variable and some set of other variable values (e.g., without the restriction of the expression being positive, with $b = c = 1$, the partial wrt $a$ would always be negative), then $a + b + c$ has no maximum. Otherwise, for the smallest derivative being positive, when you increase that variable's value by $1$, the function increases less than if either of the $2$ other variables increased by $1$. This means for the next larger value of $a + b + c$, the function value will increase the least, so more increases in values of that smallest derivative variable can occur than either of the other $2$ before the expression reaches its maximum value. As such, the largest value of $a + b + c$ will be reached with that smallest derivative variable being maximized.
If there's also a strict ordering among the other $2$ variables' derivatives, then maximize the next smallest derivative variable and, finally, maximize the last variable. Otherwise, even if the other $2$ variables derivatives' are not always in the same relative order, we can at least simplify the expression we only need to determine the maximum among just the $2$ remaining variables.
To check your specific expression, first multiply out the terms and define
$$f(a, b, c) = 2^{a + b + c + 1} - 2^{a + b + 1} - 3\left(2^{a + 1}\right) \tag{1}\label{eq1A}$$
Next, note if $d$ denotes an expression composed only of constants, $b$ and/or $c$, then since $2^{a + d} = e^{\ln(2)(a + d)}$, the chain rule gives $\frac{\partial(2^{a + d})}{\partial a} = \ln(2)e^{\ln(2)(a + d)} = \ln(2)2^{a + d}$. We thus get
$$\frac{\partial f(a, b, c)}{\partial a} = \ln(2)\left(2^{a + b + c + 1} - 2^{a + b + 1} - 3\left(2^{a + 1}\right)\right) \tag{2}\label{eq2A}$$
$$\frac{\partial f(a, b, c)}{\partial b} = \ln(2)\left(2^{a + b + c + 1} - 2^{a + b + 1}\right) \tag{3}\label{eq3A}$$
$$\frac{\partial f(a, b, c)}{\partial c} = \ln(2)\left(2^{a + b + c + 1}\right) \tag{4}\label{eq4A}$$
Fortunately, the ordering $\frac{\partial f(a, b, c)}{\partial c} \gt \frac{\partial f(a, b, c)}{\partial b} \gt \frac{\partial f(a, b, c)}{\partial a} \gt 0$ always holds. This means the variable to maximize first is $a$. This maximum occurs when $2^b(2^c-1) - 3$ is the smallest positive integer. In this case, it's $1$, with $b = 2$ and $c = 1$ being the only valid values. We thus get $a + 1 = \left\lfloor \log_2(125) \right\rfloor \implies a = 5$.
This gives
$$a + b + c \le 1 + 2 + 5 = 8 \tag{5}\label{eq5A}$$
so $K = 9$ is the smallest integer where $a + b + c \lt K$ is always true. In contrast, you've shown that $c$'s maximum is $4$, but only $a = b = 1$ works then, with this giving $a + b + c = 6$.
Note this process is related to, but is more general than, what InterstellarProble wrote, starting at this question comment. Nonetheless, as I stated earlier, this method will not be sufficient in many cases where the partial derivatives don't always have the same relative ordering for all of the domain values.