Given is equivalent to $\displaystyle \sum \dfrac{a\dfrac{a+b+c}{3}+\dfrac{(a+b+c)^2}{3}}{a(a+b+c)+bc}\geq 3\Leftrightarrow \sum \dfrac{(a+b+c)(a+b+a+c)}{(a+b)(a+c)}\geq 9$. Set $a+b=2x,b+c=2y,c+a=2z$.We have to show that $\displaystyle \sum \dfrac{(x+y+z)(2x+2z)}{2x2z}\geq 9\Leftrightarrow \sum\dfrac{x+z}{xz}\geq \dfrac{18}{x+y+z}$ which is true because $\displaystyle \sum\dfrac{x+z}{xz}=2\sum \dfrac{1}{x}\geq 2\dfrac{(1+1+1)^3}{x+y+z}=\dfrac{18}{x+y+z}$QED

parmenides51 wrote: Let $a,b,c > 0$ be real numbers with the sum equal to $3$. Show that: $$\frac{a+3}{3a+bc}+\frac{b+3}{3b+ca}+\frac{c+3}{3c+ab} \ge 3$$ https://artofproblemsolving.com/community/c6h1902272p13007986 https://artofproblemsolving.com/community/c6h275851p1492806 $$ LHS = \sum \frac {2a + b + c}{(a + b)(a + c)}= \sum \frac {(a + b) + (a + c)}{(a + b)(a + c)} = 2\sum \frac {1}{a + b}\ge \frac {18}{2(a + b + c)} = 3$$

Let $a,b,c > 0$ and $a+b+c=3.$ Show that: $$\frac{a+1}{3a+bc}+\frac{b+1}{3b+ca}+\frac{c+1}{3c+ab} \ge \frac{3}{2}$$$$\frac{a+2}{3a+bc}+\frac{b+2}{3b+ca}+\frac{c+2}{3c+ab} \ge \frac{9}{4}$$