= a( 1-x)/x+bx/( 1-x)+(a+b)

Because 0

And a and b are also normal numbers, so according to the basic inequality:

a( 1-x)/x+bx/( 1-x)>= 2 √( a( 1-x)/x * bx/( 1-x))= 2√(ab)

that is

f(x)>=2√ab+a+b

At this time, a( 1-x)/x=bx/( 1-x).

X=√a/(√a+√b)。

The minimum value of f(x) is 2 √ ab+a+b.