Nice Inequality problem.
 

Let $a,b,c>0$ with $abc=1$

Prove that $$\displaystyle 2\Big(\Big(\frac{a}{1+a+ab}\Big)^2+\Big(\frac{b}{1+b+bc}\Big)^2+\Big(\frac{c}{1+c+ca}\Big)^2\Big)+\frac{9}{(1+a+ab)(1+b+bc)(1+c+ ca)}\ge 1$$

ขอคำใบ้หน่อยได้มั้ยครับ คิดมานานละยังไม่ออกเลย เเหะๆ

$abc=1$ จะได้ว่า $\displaystyle (1+a+ab)(1+b+bc)(1+c+ca)=\sum_{cyc} a(1+b+bc)(1+c+ca)$ ครับ

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.