problem with comparing inequality Hello I just wondering if it's possible to prove this inequality: there are positive, various $a,b,c$ and 
$ \frac{3a-b}{3} \ge x \ge \frac{3(a^2-b^2)}{3a+b}$
$ \frac{3b-c}{3} \ge y \ge \frac{3(b^2-c^2)}{3b+c}$
$ \frac{3c-a}{3} \ge z \ge \frac{3(c^2-a^2)}{3c+a}$
I want to prove that $x+y \ge z$ when we add we get 
$\frac{3a+2b-c}{3} \ge x+y \ge \frac {3(a-b)(a+b)}{3 a+b}+\frac {3(b-c)(b+c)}{3 b+c} $ but further I don't know how to compare it with $z$

Hint:
To prove that $x+y\ge z$ we need to prove that minimum value of $x$ plus minimum value of $y$ is $\ge$ maximum value of $z$.
So you have:
$x=\frac{3(a^2-b^2)}{3a+b}; y=\frac{3(b^2-c^2)}{3b+c}; z=\frac{3c-a}{3}$
Just prove that $\frac{3(a^2-b^2)}{3a+b}+\frac{3(b^2-c^2)}{3b+c}\ge\frac{3c-a}{3}$
