59th Ukrainian National Mathematical Olympiad

Show that for positive numbers $x,y,z,a,b,c$ so that $x+y+z=a+b+c$ the following inequality holds: $$\frac{x}{a+b}+\frac{y}{b+c}+\frac{z}{c+a}+\frac{a}{x+z}+\frac{b}{x+y}+\frac{c}{y+z}>2.$$