Let $x = \frac{a}{1+a+ab}, y = \frac{b}{1+b+bc}, z = \frac{c}{1+c+ca}$
From the above hint, we get $x+y+z = 1$.
It remains to show that $2(x^2+y^2+z^2)+9xyz \geq 1$
$\leftrightarrow 2(x^2+y^2+z^2)(x+y+z)+9xyz \geq (x+y+z)^3$
$\leftrightarrow x^3+y^3+z^3+3xyz \geq x^2y+x^2z+y^2x+y^2z+z^2x+z^2y$
which is true by Schur's inequality.
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