Let \(x,y,z,\) be positive real numbers. Find all possible values of \(f(x,y,z) = \frac{x}{x + y} + \frac{y}{y + z} + \frac{z}{z + x}\)

Guest Mar 11, 2019

#1**+1 **

\(f(x,y,z) = \dfrac{1}{1+\dfrac{y}{x}} + \dfrac{1}{1+\dfrac{z}{y}} +\dfrac{1}{1+\dfrac{x}{z}}\)

Let y/x = a, z/y = b, x/z = c. That means abc = 1 and a,b,c are positive real numbers.

\(f(a,b,c) = \dfrac{1}{1+a} + \dfrac{1}{1+b} + \dfrac{1}{1+c}\\ f(a,b,c) = \dfrac{(1+a+b+ab)+(1+b+c+bc)+(1+a+c+ac)}{(1+a+ab+b)(1+c)}\\ f(a,b,c) = \dfrac{3+2(a+b+c)+(ab+bc+ca)}{1+ab+bc+ca+a+b+c+abc}\\ f(a,b,c) = \dfrac{3+2(a+b+c)+(ab+bc+ca)}{2 + (a+b+c) + (ab+bc+ca)}\\ f(a,b,c) = 1 + \dfrac{1 + a + b + c}{2 + a + b + c + ab + bc + ca}\)

Minimum is attained when a = b = c = 1.

min f = 1 + 1/2 = 3/2.

\(\therefore f(x,y,z) \geq \dfrac{3}{2}\) <- this is the range of values of f(x,y,z).

MaxWong Mar 12, 2019