ta có $\frac{1}{4}(a+b+c)^{3} \leq (a+b)^{3}+c^{3} \leq 4(a^{3}+b^{3})+c^{3} \leq 2(a+b+c)(\frac{(a+b+c)^{2}}{4}-2$$\Rightarrow a+b+c\geq 4$
$\frac{2a^{2}}{3a^{2}+b^{2}+2a(c+2)}=\frac{a}{a+c+2+\frac{b^{2}}{2a}+\frac{a}{2}}\leq \frac{a}{a+c+2+2\sqrt{\frac{b^{2}}{2a}\frac{a}{2}}}=\frac{a}{a+b+c+2}$
có $(a+b)^{2}+c^{2} \geq \frac{1}{2}(a+b+c)^{2}$
khi đó P $\leq \frac{a+b+c}{a+b+c+2}-\frac{(a+b+c)^{2}}{32}$ Đặt $t=a+b+c \geq4$
$\Rightarrow P \leq \frac{t}{t+2} -\frac{t^{2}}{32}=f(t)$
$f'(t)=\frac{2}{(t+2)^{2}}-\frac{t}{16}<0 ,\forall t\geq4$
$\Rightarrow f(t) nb/ \left[ 4{;} +vô cùng\right]$
$\Rightarrow P \leq f(t) \leq f(4)=\frac{1}{6}$
dấu "=" $\Leftrightarrow a=b=1;c=2$