28 tháng 11 2020 lúc 20:20
\(A\ge\sqrt{2x+1}+\sqrt{2y+1}+\sqrt{2z+1}\)
Do \(0\le x;y;z\le4\Rightarrow1\le\sqrt{2x+1};\sqrt{2y+1};\sqrt{2z+1}\le3\)
Đặt \(\left(\sqrt{2x+1};\sqrt{2y+1};\sqrt{2z+1}\right)=\left(a;b;c\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a^2+b^2+c^2=11\\1\le a;b;c\le3\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}\left(a-1\right)\left(a-3\right)\le0\\\left(b-1\right)\left(b-3\right)\le0\\\left(c-1\right)\left(c-3\right)\le0\end{matrix}\right.\) \(\Rightarrow a+b+c\ge\frac{a^2+b^2+c^2+9}{4}=5\)
\(A_{min}=5\) khi \(\left(x;y;z\right)=\left(4;0;0\right)\)
