ĐKXĐ:...
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}+\sqrt{z}=12\\\frac{\sqrt{x}}{5}+\frac{\sqrt{y}}{2}+\sqrt{z}=\frac{\sqrt{x}}{5}.\frac{\sqrt{y}}{2}.\sqrt{z}\end{matrix}\right.\)
Đặt \(\left(\frac{\sqrt{x}}{5};\frac{\sqrt{y}}{4};\frac{\sqrt{z}}{3}\right)=\left(a;b;c\right)\)
\(\Rightarrow\left\{{}\begin{matrix}5a+4b+3c=12\\a+2b+3c=6abc\end{matrix}\right.\)
Từ pt đầu ta có:
\(12=5a+4b+3c\ge12\sqrt[12]{a^5.b^4.c^3}\Leftrightarrow a^5b^4c^3\le1\) (1)
Từ pt sau:
\(6abc=a+2b+3c\ge6\sqrt[6]{ab^2c^3}\Leftrightarrow abc\ge\sqrt[6]{ab^2c^3}\)
\(\Leftrightarrow a^6b^6c^6\ge ab^2c^3\Leftrightarrow a^5b^4c^3\ge1\) (2)
Từ (1) và (2) \(\Rightarrow a^5b^4c^3=1\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=1\)
\(\Rightarrow\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)=\left(5;4;3\right)\Rightarrow\left(x;y;z\right)=\left(25;16;9\right)\)
