Prove that for $a, b, c > 0$ where $a + b + c = 6$, the following inequality holds: $$ \frac{a}{a^2+bc+4} + \frac{b}{b^2+ca+4} + \frac{c}{c^2+ab+4} \leq \frac{1}{2}. $$
From the AM-GM inequality, we find that $ abc \leq 8 $. I attempted to express the first denominator as $(6-b-c)^2+bc+4 = (b+c)^2 - 12(b+c) + bc + 40$. Substituting every $a$ in the inequality with $6-b-c$, the calculations seemed too complicated. I also explored exploiting monotonicity after achieving a common denominator for each fraction.
Proof.
By AM-GM $a^2+4\ge 4a.$ It implies $$a^2+bc+4\ge 4a+bc \iff \frac{a}{a^2+bc+4}\le \frac{a}{4a+bc}.$$
And we'll prove stronger inequality$$\frac{a}{4a+bc}+\frac{b}{4b+ca}+\frac{c}{4c+ab}\le \frac{1}{2}$$ $$\iff \frac{4a+bc-bc}{4a+bc}+\frac{4b+ca-ca}{4b+ca}+\frac{4c+ab-ab}{4c+ab}\le 2 $$ or$$\frac{bc}{4a+bc}+\frac{ca}{4b+ca}+\frac{ab}{4c+ab}\ge 1.$$ Notice. Equality holds iff $a=b=c=2,$ also for $a=b=3,c=0$ and its permutations.
Now, using Cauchy-Schwarz inequality$$\sum_{\mathrm{cyc}}\frac{bc}{4a+bc}\ge \frac{(ab+bc+ca)^2}{12abc+(bc)^2+(ca)^2+(ab)^2}=\frac{(ab+bc+ca)^2}{2abc(a+b+c)+(bc)^2+(ca)^2+(ab)^2}= 1.$$ Hence, the proof is done. Equality holds iff $a=b=c=2.$