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\(A=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-3}}{z}\)
Áp dụng BĐT AM-GM ta có:
\(A\le\frac{1+x-1}{x}+\frac{2+y-2}{2y}+\frac{3+z-3}{3z}=1+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}=1\\\sqrt{y-2}=2\\\sqrt{z-3}=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x-1=1\\y-2=2\\z-3=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=4\\z=6\end{cases}}\)
Vậy \(A_{max}=\frac{11}{6}\Leftrightarrow\hept{\begin{cases}x=2\\y=4\\z=6\end{cases}}\)
Xin lỗi bạn. Bài đó mk lm sai rồi.
Sửa:
Áp dụng BĐT AM-GM ta có:
\(A=\frac{1.\sqrt{x-1}}{x}+\frac{\sqrt{2}.\sqrt{y-2}}{\sqrt{2}.y}+\frac{\sqrt{3}.\sqrt{z-3}}{\sqrt{3}.z}\le\frac{\frac{1+x-1}{2}}{x}+\frac{\frac{2+y-2}{2}}{\sqrt{2}.y}+\frac{\frac{3+z-3}{2}}{\sqrt{3}.z}=\frac{1}{2}+\frac{1}{2.\sqrt{2}}+\frac{1}{2.\sqrt{3}}\)\(=\frac{\sqrt{6}+\sqrt{3}+\sqrt{2}}{2.\sqrt{6}}\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}=1\\\sqrt{y-2}=\sqrt{2}\\\sqrt{z-3}=\sqrt{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}x-1=1\\y-2=2\\z-3=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=4\\z=6\end{cases}}\)
Vậy \(A_{max}=\frac{\sqrt{6}+\sqrt{2}+\sqrt{3}}{2.\sqrt{6}}\)\(\Leftrightarrow\hept{\begin{cases}x=2\\y=4\\z=6\end{cases}}\)
\(\Leftrightarrow2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=x+y+z\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-1-2\sqrt{y-1}+1\right)+\left(z-2-2\sqrt{z-2}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x}-1=0\\\sqrt{y-1}-1=0\Leftrightarrow\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\\\sqrt{z-2}-1=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x}-1=0\\\sqrt{y-1}-1=0\Leftrightarrow\\\sqrt{z-2}-1=0\end{cases}\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}}\)
vậy \(S=x+y=1+2=3\)
\(\frac{xy\sqrt{z-1}+xz\sqrt{y-2}+yz\sqrt{x-3}}{xyz}\\ =\frac{xy\sqrt{z-1}}{xyz}+\frac{xz\sqrt{y-2}}{xyz}+\frac{yz\sqrt{x-3}}{xyz}\\ =\frac{\sqrt{z-1}}{z}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{x-3}}{x}\\ =\frac{2\sqrt{z-1}}{2z}+\frac{2\sqrt{2}\sqrt{y-2}}{2\sqrt{2}y}+\frac{2\sqrt{3}\sqrt{x-3}}{2\sqrt{3}x}\)
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.......
chuyển vế rồi thêm bớt cậu sẽ có rồi tìm được x=1 y=1 z=4
\(\left(x-2\sqrt{x}+1\right)+\left(y-2\sqrt{y}+1\right)+\left(z-4\sqrt{z}+4\right)=0\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-1\right)^2+\left(\sqrt{z}-2\right)^2=0\)
Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\Leftrightarrow\left(x+y\right)\left(\frac{zx+z^2+zy+xy}{xyz\left(x+y+z\right)}\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left[z\left(x+z\right)+y\left(x+z\right)\right]=0\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Rightarrow\left(x^2-y^2\right)\left(y^3+z^3\right)\left(z^4-x^4\right)=0\).
Vậy \(M=\frac{3}{4}+\left(x^2-y^2\right)\left(y^3+z^3\right)\left(z^4-x^4\right)=\frac{3}{4}+0=\frac{3}{4}\)
Áp dụng BĐT Cô-si dạng Engel,ta có :
\(\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{xz}}+\frac{z^2}{z+\sqrt{xy}}\ge\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}\)
Mà \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le x+y+z\)
\(\Rightarrow\)\(\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\ge\frac{3}{2}\)
Dấu "=" xảy ra khi x = y = z = \(\frac{3}{2}\)
làm tương trự như bài trên nhá