Inequality: $ \frac{a}{1+9bc+k(b-c)^2}+\frac{b}{1+9ca+k(c-a)^2}+\frac{c}{1+9ab+k(a-b)^2}\geq\frac{1}{2} $ I was trying to solve this inequality, but I wasn't able to do so:
Find the maximum number $k\in\mathbb R$ such that:
$$ \frac{a}{1+9bc+k(b-c)^2}+\frac{b}{1+9ca+k(c-a)^2}+\frac{c}{1+9ab+k(a-b)^2}\geq\frac{1}{2} $$
Holds for all $a,b,c\ge0$ with $a+b+c=1$
Any help is highly appreciated.

Plugging, $a=1/2,b=1/2,c=0 \implies k \le 4$.
I will show that $k=4$. 
$$ \frac{a}{1+9bc+4(b-c)^2}+\frac{b}{1+9ca+4(c-a)^2}+\frac{c}{1+9ab+4(a-b)^2}\ge \frac{1}{2} \\ \Longleftrightarrow \frac{a^2}{a+9abc+4a(b-c)^2}+\frac{b^2}{b+9abc+4b(c-a)^2}+\frac{c^2}{c+9abc+4c(a-b)^2} \ge \frac{1}{2}$$
By cauchy it suffice to show that, $$  \frac{(a+b+c)^2}{a+b+c+3abc+4(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)} \ge \frac{1}{2} \\ \Longleftrightarrow  \frac{1}{1+3abc+4(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)} \ge \frac{1}{2} \\ \Longleftrightarrow 1 \ge 3abc+4a^2b+b^2c+c^2a+ab^2+bc^2+ca^2  \\  \Longleftrightarrow (a+b+c)^3 \ge 3abc+a^2b+b^2c+c^2a+ab^2+bc^2+ca^2 \\ \Longleftrightarrow a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b) \ge 0 $$ 
Last inequality is true by Schur. $\Box$
