$x,y,z\ge 0$, $x+y+z=2$, prove that $$ \frac{x}{1+y^2}+\frac{y}{1+z^2}+\frac{z}{1+x^2}\ge\frac{18}{13} $$
I had no idea besides brute-force: multiply everything, dehomogenize and use Muirhead. But after a partial estimate I got a sufficient inequality, which is not true, so it wasn't the way.
We need to prove that $$\sum_{cyc}\frac{x}{\frac{(x+y+z)^2}{4}+y^2}\geq\frac{36}{13(x+y+z)}$$ or $$\sum_{cyc}(5x^6+22x^5y+9x^5z-6x^4y^2+20x^4z^2-7x^3y^3+39x^4yz-26x^3y^2z+13x^3z^2y-69x^2y^2z^2)\geq0,$$ which is true by AM-GM and Muirhead.
Indeed, by Muirhead $$\sum_{cyc}(5x^6+9x^5y+9x^5z-6x^4y^2-6x^4z^2)\geq11\sum_{cyc}x^3y^3.$$ Thus, it's remains to prove that $$\sum_{cyc}(13x^5y+26x^4z^2+4x^3y^3+39x^4yz-26x^3y^2z+13x^3z^2y-69x^2y^2z^2)\geq0.$$ Now, by AM-GM $$\sum_{cyc}x^4z^2=\frac{1}{2}\sum_{cyc}(x^4z^2+y^4x^2)\geq\sum_{cyc}x^3y^2z$$ and the rest is AM-GM again.
For example, $$\sum_{cyc}x^5y\geq3\sqrt[3]{x^5y\cdot y^5z\cdot z^5x}=3x^2y^2z^2.$$
Also, we can use the following way.
After homogenization in the my first proof we can assume that $x+y+z=3$.
Thus, we need to prove that $$\sum_{cyc}\frac{x}{9+4y^2}\geq\frac{3}{13}.$$ Now, we see that $$\frac{1}{9+4x^2}\geq-\frac{20}{1521}x^2-\frac{32}{1521}x+\frac{1}{9}$$ it's $$x(5x+18)(x-1)^2\geq0.$$ Thus, it's enough to prove that $$\sum_{cyc}x\left(-\frac{20}{1521}y^2-\frac{32}{1521}y+\frac{1}{9}\right)\geq\frac{3}{13}$$ or $$\sum_{cyc}(8xy+5x^2z)\leq39.$$ Now, let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.
Thus, by Rearrangement and AM-GM we obtain: $$x^2z+y^2x+z^2y=x\cdot xz+y\cdot yx+z\cdot zy\leq a\cdot ab+b\cdot ac+c\cdot bc=$$ $$=b(a^2+ac+c^2)=b(a+c)^2-abc=$$ $$=4b\left(\frac{a+c}{2}\right)^2-xyz\leq4\left(\frac{b+\frac{a+c}{2}+\frac{a+c}{2}}{3}\right)^3-xyz=4-xyz.$$ Thus, it remains to prove that $$8(xy+xz+yz)-5xyz\leq19.$$ Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that $$24uv^2-5w^3\leq19u^3.$$ But by Schur $$w^3\geq4uv^2-3u^3.$$
Thus, it's enough to prove that $$19u^3+5(4uv^2-3u^3)-24uv^2\geq0$$ or $$u(u^2-v^2)\geq0,$$ which is obvious.
Done!