Đề này hình như nhầm chỗ căn,phải là 2 chữ c,không có b
$a^{2}+b^{2}+c^{2}+4\geq \frac{(a+b+c+2)^{2}}{4}\Rightarrow \sqrt{a^{2}+b^{2}+c^{2}+4}\geq \frac{1}{2}(a+b+c+2)$
Áp dụng BĐT AM-GM ta có:
$(a+b)\sqrt{(a+2c)(a+2c)}\leq (a+b)\frac{1}{2}(a+2c+a+2c)=\frac{1}{6}(3a+3b)(a+b+4c)\leq \frac{(4a+4b+4c)^{2}}{24}=\frac{2}{3}(a+b+c)^{2}$
$\Rightarrow P\leq \frac{8}{a+b+c+2}-\frac{27}{2(a+b+c)^{2}}$
Đặt $t=a+b+c$
$P\leq \frac{8}{t+2}-\frac{27}{2t^{2}}$
Chọn điểm rơi là t=6
$P\leq \frac{8}{\frac{2t}{3}+\left ( \frac{t}{3} +2\right )}-\frac{27}{2t^{2}}\leq \frac{8}{4}\left ( \frac{1}{\frac{2t}{3}} +\frac{1}{\frac{t}{3}+2}\right )-\frac{27}{2t^{2}}\leq 2\left [ \frac{3}{2t} +\frac{1}{4}\left ( \frac{3}{t}+\frac{1}{2} \right )\right ]-\frac{27}{2t^{2}}=\frac{9}{2t}+\frac{1}{4}-\frac{27}{2t^{2}}=\frac{5}{8}-\frac{3}{2}(\frac{3}{t}-\frac{1}{2})^{2}\leq \frac{5}{8}$