\(a+b+c=3\Rightarrow0\le a;b;c\le3\)
Đặt \(\left\{{}\begin{matrix}\sqrt{3+a}=x\\\sqrt{3+b}=y\\\sqrt{3+c}=z\end{matrix}\right.\) \(\Rightarrow\sqrt{3}\le x;y;z\le\sqrt{6}\)
\(x^2+y^2+z^2=a+b+c+9=12\)
\(\left(x-\sqrt{3}\right)\left(x-\sqrt{6}\right)\le0\Rightarrow x^2-\left(\sqrt{3}+\sqrt{6}\right)x+3\sqrt{2}\le0\)
\(\Rightarrow x\ge\frac{x^2+3\sqrt{2}}{\sqrt{3}+\sqrt{6}}\)
Tương tự ta có \(y\ge\frac{y^2+3\sqrt{2}}{\sqrt{3}+\sqrt{6}}\); \(z\ge\frac{z^2+3\sqrt{2}}{\sqrt{3}+\sqrt{6}}\)
\(\Rightarrow P=x+y+z\ge\frac{x^2+y^2+z^2+9\sqrt{2}}{\sqrt{3}+\sqrt{6}}=\frac{12+9\sqrt{2}}{\sqrt{3}+\sqrt{6}}=\sqrt{6}+2\sqrt{3}\)
\(\Rightarrow P_{min}=\sqrt{6}+2\sqrt{3}\) khi \(\left(a;b;c\right)=\left(3;0;0\right)\) và các hoán vị
