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ta có : \(C=\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+2}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{2\sqrt{z}}{\sqrt{xyz}+\sqrt{xz}+2\sqrt{z}}\)
\(=\dfrac{\sqrt{x}}{\sqrt{xyz}+\sqrt{xy}+\sqrt{x}}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{2}{\sqrt{xy}+\sqrt{x}+2}\)
\(=\dfrac{1}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{xyz}}{\sqrt{xyz}+\sqrt{xy}+\sqrt{x}}\)
\(=\dfrac{\sqrt{y}+1}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{yz}}{\sqrt{yz}+\sqrt{y}+1}=\dfrac{\sqrt{yz}+\sqrt{y}+1}{\sqrt{yz}+\sqrt{y}+1}=1\)
\(P=\frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+2}+\frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\frac{2\sqrt{z}}{\sqrt{xz}+2\sqrt{x}+2}\)
\(=\frac{\sqrt{xz}}{\sqrt{xyz}+\sqrt{xz}+2\sqrt{z}}+\frac{\sqrt{xyz}}{\sqrt{xyz^2}+\sqrt{xyz}+\sqrt{xz}}+\frac{2\sqrt{x}}{\sqrt{xz}+2\sqrt{z}+2}\)
\(=\frac{\sqrt{xz}}{\sqrt{xz}+2\sqrt{x}+2}+\frac{2}{2\sqrt{x}+2+\sqrt{xz}}+\frac{2\sqrt{z}}{\sqrt{xz}+2\sqrt{z}+2}\) (do \(xyz=4\))
\(=\frac{\sqrt{xz}+2\sqrt{z}+2}{\sqrt{xz}+2\sqrt{z}+2}=1\)
Ta có: \(xyz=4\Rightarrow\sqrt{xyz}=2\)
Thay vào biểu thức P thì được:
\(P=\frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+\sqrt{xyz}}+\frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\frac{\sqrt{xyz^2}}{\sqrt{zx}+\sqrt{xyz^2}+\sqrt{xyz}}\)
\(P=\frac{1}{\sqrt{yz}+\sqrt{y}+1}+\frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\frac{\sqrt{yz}}{\sqrt{yz}+\sqrt{y}+1}\)
\(P=\frac{1+\sqrt{y}+\sqrt{yz}}{\sqrt{yz}+\sqrt{y}+1}=1\Rightarrow\sqrt{P}=1.\)
Vậy ...
\(P=\frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+1}+\frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\frac{\sqrt{z}}{\sqrt{xz}+\sqrt{z}+1}\)( Vì xyz=1 nên \(\sqrt{xyz}=1\))
\(P=\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{y}+1+\sqrt{yz}\right)}+\frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\frac{\sqrt{z}}{\sqrt{z}\left(\sqrt{x}+1+\sqrt{xy}\right)}\)
\(P=\frac{\sqrt{y}+1}{\sqrt{y}+1+\sqrt{yz}}+\frac{1}{\sqrt{x}+1+\sqrt{xy}}\)
\(P=\frac{\sqrt{y}+1}{\sqrt{y}+1+\sqrt{yz}}+\frac{\sqrt{xyz}}{\sqrt{x}\left(1+\sqrt{yz}+\sqrt{y}\right)}\)
\(P=\frac{\sqrt{y}+1}{\sqrt{y}+1+\sqrt{yz}}+\frac{\sqrt{yz}}{\sqrt{y}+1+\sqrt{yz}}=\frac{\sqrt{y}+1+\sqrt{yz}}{\sqrt{y}+1+\sqrt{yz}}=1\)
\(\frac{xy\sqrt{z-1}+xz\sqrt{y-2}+yz\sqrt{x-3}}{xyz}\\ =\frac{xy\sqrt{z-1}}{xyz}+\frac{xz\sqrt{y-2}}{xyz}+\frac{yz\sqrt{x-3}}{xyz}\\ =\frac{\sqrt{z-1}}{z}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{x-3}}{x}\\ =\frac{2\sqrt{z-1}}{2z}+\frac{2\sqrt{2}\sqrt{y-2}}{2\sqrt{2}y}+\frac{2\sqrt{3}\sqrt{x-3}}{2\sqrt{3}x}\)
Áp dụng BDT Cô-si với 2 số không âm:
\(\Rightarrow\frac{2\sqrt{z-1}}{2z}+\frac{2\sqrt{2}\sqrt{y-2}}{2\sqrt{2}y}+\frac{2\sqrt{3}\sqrt{x-3}}{2\sqrt{3}x}\\ \le\frac{1+\left(z-1\right)}{2z}+\frac{2+\left(y-2\right)}{2\sqrt{2}y}+\frac{3+\left(x-3\right)}{2\sqrt{3}x}\\ =\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}=\frac{1}{2}+\frac{\sqrt{2}}{4}+\frac{\sqrt{3}}{6}\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}z-1=1\\y-2=2\\x-3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=2\\y=4\\x=6\end{matrix}\right.\)
Vậy.......
Bạn tham khảo lời giải tại đây:
Câu hỏi của Angela jolie - Toán lớp 9 | Học trực tuyến
\(3-2P=\frac{x}{x+2\sqrt{yz}}+\frac{y}{y+2\sqrt{xz}}+\frac{z}{z+2\sqrt{xy}}\)
\(3-2P\ge\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}=1\)
\(\Rightarrow2P\le2\Rightarrow P\le1\)
Dấu "=" xảy ra khi \(x=y=z\)
\(M\le\sqrt{\left(1+1\right)\left(x+y+2\right)}=\sqrt{20}=4\sqrt{5}\)
\(M_{max}=4\sqrt{5}\) khi \(\left\{{}\begin{matrix}x-2=y+4\\x+y=8\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)
