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 : $\displaystyle x+y+z> 1$
Without the loss of generality we can assume
a >= b >= c > 0
Under this what can you say about x,y,z ?
Originally Posted by 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 : $\displaystyle x+y+z> 1$