Do \(\left\{{}\begin{matrix}x;y;z\ge0\\x+y+z=1\end{matrix}\right.\) \(\Rightarrow0\le x;y;z\le1\) \(\Rightarrow\left\{{}\begin{matrix}1\le\sqrt{3x+1}\le2\\1\le\sqrt{3y+1}\le2\\1\le\sqrt{3z+1}\le2\end{matrix}\right.\)
Đặt \(\left(\sqrt{3x+1};\sqrt{3y+1};\sqrt{3z+1}\right)=\left(a;b;c\right)\)
\(\Rightarrow\left\{{}\begin{matrix}1\le a;b;c\le2\\a^2+b^2+c^2=3\left(x+y+z\right)+3=6\end{matrix}\right.\)
Khi đó \(A=a+b+c\)
Do \(1\le a\le2\Rightarrow\left(a-1\right)\left(a-2\right)\le0\)
\(\Leftrightarrow a^2-3a+2\le0\)
\(\Leftrightarrow a^2+2\le3a\Leftrightarrow a\ge\frac{a^2+2}{3}\)
Tương tự ta có: \(b\ge\frac{b^2+2}{3}\) ; \(c\ge\frac{c^2+2}{3}\)
Cộng vế với vế: \(a+b+c\ge\frac{a^2+b^2+6}{3}=\frac{6+6}{3}=4\)
\(A_{min}=4\) khi \(\left(a;b;c\right)=\left(1;1;2\right)\) và hoán vị hay \(\left(x;y;z\right)=\left(0;0;1\right)\) và hoán vị
