# Prove that

#### dhiab

If a , b and c are three positf real numbers and :
$$\displaystyle x=\frac{a}{a+b}$$
$$\displaystyle y=\frac{b}{b+c}$$
$$\displaystyle z=\frac{c}{c+a}$$
Prove that : $$\displaystyle x+y+z> 1$$

#### pickslides

MHF Helper
$$\displaystyle \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}=\dots$$ ??

#### aman_cc

Without the loss of generality we can assume
a >= b >= c > 0

Under this what can you say about x,y,z ?

If a , b and c are three positf real numbers and :
$$\displaystyle x=\frac{a}{a+b}$$
$$\displaystyle y=\frac{b}{b+c}$$
$$\displaystyle z=\frac{c}{c+a}$$
Prove that : $$\displaystyle x+y+z> 1$$

#### roninpro

It may also help to scale the numbers so that $$\displaystyle a+b+c=1$$ and sharpen the lower bound to $$\displaystyle x+y+z\geq \frac{3}{2}$$.