Có $14+2bc=a^2+(b+c)^2\geq 2a(b+c)\Rightarrow a(b+c)\leq7+bc$$\Rightarrow \frac{4a}{a^2+bc+7}\leq \frac{4}{a+b+c}$
$a(b+c)\leq (\frac{a+b+c}{2})^2$
$\Rightarrow -\frac{3}{a(b+c)}\leq -\frac{12}{(a+b+c)^2}$
$\frac{4}{a+b+c}-\frac{12}{(a+b+c)^2} \leq \frac{1}{3}\Rightarrow \frac{4a}{a^2+bc+7}-\frac{3}{a^2+bc+7}\leq \frac{1}{3}$
$(1+\frac{1}{3})(a^2+3c^2) \geq (a+b)^2\Rightarrow a^2+3c^2\geq \frac{3}{4}(a+b)^2$
$\Rightarrow \frac{4(a+c)}{a^2+3c^2+28}\leq \frac{4(a+c)}{\frac{3}{4}(a+c)^2+28}\leq \frac{1}{25}(a+c-4)+\frac{2}{5}$
$-\frac{5}{(a+b)^2}\leq \frac{2}{25}(a+b-5)-\frac{1}{5}$
$\Rightarrow \frac{4(a+c)}{a^2+3c^2+28}-\frac{5}{(a+b)^2}\leq \frac{3a+2b+c-9}{25}$
$(3a+2b+c)\leq \sqrt{(a^2+b^2+c^2)(3^2+2^2+1^2)}$
Từ đó $Max=\frac{8}{15}$