How to find the factors whose sum is minimum Lets take a number 108.
How to find natural numbers a and b such that ab=108
but there sum should be minimum.
Please show the solution for number 108.
 A: Assuming real $a,b>0$
$$a+b=a+\frac{108}a=\left(\sqrt a-\sqrt{\frac{108}a}\right)^2+2\sqrt{a\cdot\frac{108}a}\ge2\sqrt{a\cdot\frac{108}a}=2\cdot6\sqrt3$$  the equality(i.e., the minimum value of $a+b$) occurs if $\displaystyle\sqrt a-\sqrt{\frac{108}a}=0\implies a=\sqrt{108}=6\sqrt3$
A: Apply the AM-GM inequality we then have:
$$\frac{a+b}{2} \ge \sqrt{ab}$$
$$\frac{a+b}{2} \ge \sqrt{108} = 6\sqrt{3}$$
From the property of AM-GM we know that the equality will hold iff $a=b$. So we get that $a=b=6\sqrt{3} \approx 10.3923048454132638$
You can think of this like the minima of the function $f(a,b) = a+b$ under constrain $g(a,b) = ab - 108 = 0$. So from this and from the fact that the square is the rectangle with smallest perimetar for given area, we conclude that the more "square-shaped" the rectangle is the more smaller the sum $a+b$ will be. (To understand even better, let $a$ be one side of the rectangle and $b$ be the second with area $108$).
So we need to find the greates divisor of 108 smaller than $6\sqrt{3}$ and the smallest divisor greater $6\sqrt{3}$ and add them.
The first one is $12$ and the second is $9$. So the minimal sum is $21$. To get it even better write the divisors of $108$ in ascending order and you can check that the product of the first and the last one will be $108$. The product of the second and second last divisor will be $108$ and so on.
