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 : $\displaystyle x+y+z> 1$

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- May 20th 2010, 11:46 PMdhiabProve that
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 : $\displaystyle x+y+z> 1$ - May 21st 2010, 12:21 AMpickslides
$\displaystyle

\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}=\dots

$ ?? - May 21st 2010, 12:47 AMaman_cc
- May 21st 2010, 05:48 AMroninpro
It may also help to scale the numbers so that $\displaystyle a+b+c=1$ and sharpen the lower bound to $\displaystyle x+y+z\geq \frac{3}{2}$.