By AM-HM, $\frac{2}{a+b}+\frac{2}{a+c}\leq \frac{\frac{1}{a}+\frac{1}{b}}{2}+ \frac{\frac{1}{a}+\frac{1}{c}}{2}\implies \frac{2(2a+b+c)}{(a+b)(a+c)}\leq \frac{ab+2bc+ca}{2abc}$. $\frac{4abc}{(a+b)(a+c)}\leq \frac{ab+2bc+ca}{2a+b+c}\implies\frac{4abc}{(a+b)(a+c)}+((a+b)+(a+c))\leq \frac{ab+2bc+ca}{2a+b+c}+(a+3)$ $\implies \frac{4abc}{(a+b)(a+c)}+(a+b)+(a+c)\leq \frac{2a^2+2ab+2bc+2ca}{2a+b+c}+3=\frac{2(a+b)(a+c)}{(a+b)+(a+c)}+3$ $\implies \frac{4abc}{(a+b)(a+c)}+\left((a+b)+(a+c)-\frac{2(a+b)(a+c)}{(a+b)+(a+c)}\right)\leq3\implies \frac{4abc}{(a+b)(a+c)}+\frac{(a+b)^2+(a+c)^2}{(a+b)+(a+c)}\leq 3$ By AM-GM, $a^3+a^3+1\geq 3a^2\implies \sum_{cyc}(2a^3-3a^2)+3\geq 0\quad (*)$ By C.S., $(1+1+1)(a^2+b^2+c^2)\geq (a+b+c)^2\implies a^2+b^2+c^2\geq 3$. By Holder, $(a^3+b^3+c^3)(a^3+b^3+c^3)(1+1+1)\geq (a^2+b^2+c^2)^3\geq 3(a^2+b^2+c^2)^2$. $3(a^3+b^3+c^3)^2\geq 3(a^2+b^2+c^2)^2\implies a^3+b^3+c^3\geq a^2+b^2+c^2\implies \sum_{cyc}(a^3-a^2)\geq 0\quad (**)$ $(*)+(**)\implies 3(a^3+b^3+c^3)-4(a^2+b^2+c^2)+3\geq 0\implies \frac{9}{4(a^2+b^2+c^2)-3(a^3+b^3+c^3)}\geq 3 $ By AM-HM, $\sum_{cyc}\frac{1}{a^2(3b+3c-5)}=\sum_{cyc}\frac{1}{a^2(4-3a)}\geq \sum_{cyc}\frac{9}{4(a^2+b^2+c^2)-3(a^3+b^3+c^3)}\geq 3$. Thus $\frac{4abc}{(a+b)(a+c)}+\frac{(a+b)^2+(a+c)^2}{(a+b)+(a+c)}\leq3\leq\sum_{cyc}\frac{1}{a^2(3b+3c-5)}$.