$a+b+c=s$ what is the upper and lower bound for $\ln a + \ln b + \ln c$ ?
$a$ , $b$ , $c$ and $s$ all are natural numbers with out bound on them.
after that what if a,b,c has some boundaries like $ 1\le a \le k_a$ ,$ 1\le b \le k_b$ and $ 1\le c \le k_c$ ?
It can be shown using the method of Lagrange multipliers that if $a,b,c$ are real numbers, then the maximal value of $abc$ (and thus $\ln(abc)$) is given when $a=b=c=\dfrac{s}{3}$.
Thus, if $s \equiv 0 \pmod 3$, then $a = b = c = \dfrac{s}{3}$ will maximixe $\ln(abc)$.
For the case where $s$ is not divisible by $3$, however, we will still get the maximal value when $a,b,c$ are as close together as possible. So:
Let $s \equiv 1 \pmod 3$. Then $a = \lceil \dfrac{s}{3} \rceil$ and $b=c = \lfloor \dfrac{s}{3} \rfloor$ will maximize $\ln(abc)$.
Let $s \equiv 2 \pmod 3$. Then $a=b = \lceil \dfrac{s}{3} \rceil$ and $c = \lfloor \dfrac{s}{3} \rfloor$ will maximize $\ln(abc)$.
The minimal value is trickier. Let $c$ be fixed and let us try to find the minimal value of $f(a,b) = ab$, where $a+b = s-c$. Here, we see that the function $f$ can be written in one variable as $f^*(a) = a(s-z-a)$, which achieves a maximal value at $a = \dfrac{s-c}{2}$ and decreases the further $a$ is from this point. Thus, to minimize $f^*$, we take $a$ to be as far from $\dfrac{s-c}{2}$ as possible and set $x = 1$, making $b = s-c-1$, and our overall product $abc$ becomes: $$(s-c-1)c.$$
Now, when is this minimal in terms of $c$? By similar reasoning, when $c$ is as far from $\dfrac{s-1}{2}$ as possible. Thus, setting $c =1$ as well, we arrive at our minimal value for $\ln(abc)$ with $a=1, b = s-2, c = 1$ as $$\ln(s-2).$$