\(\sqrt{6\left(a^2+5\right)}=\sqrt{6\left(a^2+ab+bc+ca\right)}=\sqrt{\left(3a+3b\right)\left(2a+2c\right)}\le\frac{1}{2}\left(5a+3b+2c\right)\)
Tương tự: \(\sqrt{6\left(b^2+5\right)}\le\frac{1}{2}\left(3a+5b+2c\right)\)
\(\sqrt{c^2+5}=\sqrt{\left(a+c\right)\left(b+c\right)}\le\frac{1}{2}\left(a+b+2c\right)\)
\(\Rightarrow P\ge\frac{3a+3b+2c}{\frac{1}{2}\left(5a+3b+2c+3a+5b+2c+a+b+2c\right)}=\frac{2\left(3a+3b+2c\right)}{3\left(3a+3b+2c\right)}=\frac{2}{3}\)
Dấu "=" xảy ra khi \(a=b=\frac{c}{2}=1\)
