FINAL ROUND

Let $x=1/a$, $y=1/b$, $z=1/c$, $u=1/d$. Then $x+y+z+u=1$, $xyzu\ne 0$, and the required inequality has the form $$\frac{xy(x+y)}{x^2-xy+y^2}+\frac{yz(y+z)}{y^2-yz+z^2}+\frac{zu(z+u)}{z^2-zu+u^2}+\frac{ux(u+x)}{u^2-ux+x^2}\le 2,(1)$$ since $$\frac{a+b}{a^2-ab+b^2}=\frac{ab(b^{-1}+a^{-1})}{a^2{b}^2(b^{-2}-a^{-1}{b}^{-1}+a^{-2})}=\frac{xy(y+x)}{y^2-xy+x^2}.$$

(similar arguments apply to the remaining summands). Since for $x\ne 0$ and $y\ne 0$ we have $x^2-xy+y^2=x^2-xy+y^2/4+3y^2/4=(x-y/2{)}^2+3y^2/4>0$ and $0\le (x-y{)}^2=x^2-2xy+y^2=(x^2-xy+y^2)-xy$. So $x^2-xy+y^2\ge xy$ and $\frac{xy}{ux}\le 1$. Similarly, $y^2-yz+z^2\le 1$, $\frac{zu}{z^2-zu+u^2}\le 1$, $\frac{u^2-ux+x^2}{ux}\le 1$. Therefore $$\frac{xy(x+y)}{x^2-xy+y^2}+\frac{yz(y+z)}{y^2-yz+z^2}+\frac{zu(z+u)}{z^2-zu+u^2}+\frac{ux(u+x)}{u^2-ux+x^2}\le$$ $$\le (x+y)+(y+z)+(z+u)+(u+x)=2(x+y+z+u),$$ and since $x+y+z+u=1$, we obtain the required inequality (1).