How prove this $(xy+yz+xz)\left(\frac{xy}{z^2+1}+\frac{yz}{x^2+1}+\frac{zx}{y^2+1}\right)\le\frac{1}{10}$ 
let $x,y,z>0$ and such $x+y+z=1$, show that
  $$(xy+yz+xz)\left(\dfrac{xy}{z^2+1}+\dfrac{yz}{x^2+1}+\dfrac{zx}{y^2+1}\right)\le\dfrac{1}{10}$$

my idea:
$$\dfrac{xy}{z^2+1}=\dfrac{xy}{z^2+(x+y+z)^2}=\dfrac{xy}{2z^2+2xy+2yz+2xz+x^2+y}$$
Maybe this is old inequality,and It is said can use Cauchy-Schwarz inequality to solve it
Thank you

we can suppose that $ x=\dfrac{a}{a+b+c} , y=\dfrac{b}{a+b+c} , z=\dfrac{c}{a+b+c} $ where $ a,b,c>0 $ are three real numbers.
suppose that $ s=a+b+c $
now $ inequality \Leftrightarrow (\sum{\dfrac{ab}{s^2}})(\sum{\dfrac{(\dfrac{ab}{s^2})}{(\dfrac{c^2+s^2}{s^2})}}) \le \dfrac{1}{10} $
so we must prove that $ \dfrac{(\sum ab )}{s^2}(\sum {\dfrac{ab}{c^2+s^2}}) \le \dfrac{1}{10} $
for $ a,b,c>0 $
