Coefficents of cubic polynomial and its least root Let $x^3-(m+n+1)x^2+(m+n-3+mn)x-(m-1)(n-1)=0$, 
be a cubic polynomial with positive roots, where $m,n \ge2$ are natural nos. For fixed $m+n$, say $15$, it turns out that least root of the polynomial will be smallest in case of $m=2,n=13$ i.e the case in which difference is largest.
I checked it for many values of $m+n$ and same thing is coming out. It seems that observation is right. How to prove it? 
I noticed that coefficient of x and constant are also of largest modulus when difference between $m,n$ is largest. Does it help anyway?

Some Remarks:
There is an explicit formula to solve cubic equations  here . So you can always use it and analyze the roots in a brute-force way. I will follow a somewhat simpler method.  I will assume that : 
in the cases you are interested in you always have three (not necessarily distinct) real roots... otherwise the question isn't saying anything useful.
The method:
Let $f(x) = x^{3} + (m+ n+1) x^{2} + (mn + m + n -3) x - (m-1)(n-1)$.
The idea is we are interested in the smallest real root of $f(x)$ and this will be somewhere in the left of the smallest root of $f^{'}(x)$ (the derivative)- which is also real. So we want to minimize the roots of $f^{'}(x)$. 
This is just a simple application of Mean Value theorem.
But $f^{'}(x)$ is a quadratic function whose roots are obtained by quadratic formula. Let us use new variable : $ p = m+ n$ and $q = m-n$. 
Then we have $$f^{'}(x) = 3x^{2} - 2(p+1)x +(\frac{p^{2}-q^{2}}{4} + p -3).$$  
In the question $p$ is fixed so it is a constant. Using the quadratic formula you can check that the minimum value of the real roots are achieved when the discriminant $$D(q): 4\left ( (p+1)^{2} - 3(\frac{p^{2}-q^{2}}{4} + p -3)\right)$$ (as a function of $q$ since $p$ is fixed anyway) is maximized. 
This is what you observe in your  calculations. 
EDIT
Regarding the discussion OP mentioned in the comments here is full set of solution to the equation $x^{3} - 16x^{2} = 68x -75 = 0$ using the software  SAGE. It is free and can be used online so anyone can try it and it is wonderful!
The solutions are: 
$$ 
\left[x = -\frac{1}{2} \, {\left(\frac{1}{18} i \, \sqrt{14141} \sqrt{3} + \frac{425}{54}\right)}^{\frac{1}{3}} {\left(i \, \sqrt{3} + 1\right)} + \frac{26 i \, \sqrt{3} - 26}{9 \, {\left(\frac{1}{18} i \, \sqrt{14141} \sqrt{3} + \frac{425}{54}\right)}^{\frac{1}{3}}} + \frac{16}{3}, \;  x = -\frac{1}{2} \, {\left(\frac{1}{18} i \, \sqrt{14141} \sqrt{3} + \frac{425}{54}\right)}^{\frac{1}{3}} {\left(-i \, \sqrt{3} + 1\right)} + \frac{-26 i \, \sqrt{3} - 26}{9 \, {\left(\frac{1}{18} i \, \sqrt{14141} \sqrt{3} + \frac{425}{54}\right)}^{\frac{1}{3}}} + \frac{16}{3}, \; x = {\left(\frac{1}{18} i \, \sqrt{14141} \sqrt{3} + \frac{425}{54}\right)}^{\frac{1}{3}} + \frac{52}{9 \, {\left(\frac{1}{18} i \, \sqrt{14141} \sqrt{3} + \frac{425}{54}\right)}^{\frac{1}{3}}} + \frac{16}{3}\right]
.$$
