prove that the minimum value of (a+x)(b+x)/(c+x) , x>-c is (√a-c+√b-c)^2
62substitute x+c=t
The question becomes same as find the minimum of (t+a-c)(t+b-c)/t where t>0
= t+(a+b-2c)+(a-c)(a-b)/t
= a+b-2c + √(a-c)(a-b) { (a-c)(a-b)/t + t/(a-c)(a-b) }
now see that the expression { (a-c)(a-b)/t + t/(a-c)(a-b) } has a minimum value of 2
so the expression has a minimum value of
(a-c)+(b-c) +√(a-c)(a-b) 2
={ √(a-c) +√ (a-b) }2
341I guess its being assumed that a,b>c
We can use inequalities to obtain after letting t = c+x,
f(t) = \frac{(a-c+t)(b-c+t)}{t} = \left(\frac{a-c}{t} + 1\right)(t + (b-c)) \\\\ \ge \left(\sqrt {(a-c)} + \sqrt{(b-c}\right)^2
(By Cauchy Schwarz inequality)