x=3;y=2;z=1 

phân tích làm hàng đẳng thức bình phương

\(=>A=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{z-3}}{z}\)

áp dụng BĐT AM-GM

\(=>\sqrt{x-1}\le\dfrac{x-1+1}{2}=\dfrac{x}{2}\)

\(=>\dfrac{\sqrt{x-1}}{x}\le\dfrac{\dfrac{x}{2}}{x}=\dfrac{1}{2}\left(1\right)\)

có \(\dfrac{\sqrt{y-2}}{y}=\dfrac{\sqrt{\left(y-2\right)2}}{\sqrt{2}.y}\)

\(=>\sqrt{\left(y-2\right)2}\le\dfrac{y-2+2}{2}=\dfrac{y}{2}\)

\(=>\dfrac{\sqrt{\left(y-2\right)2}}{\sqrt{2}.y}\le\dfrac{\dfrac{y}{2}}{\sqrt{2}.y}=\dfrac{1}{2\sqrt{2}}\left(2\right)\)

tương tự \(=>\dfrac{\sqrt{z-3}}{z}\le\dfrac{1}{2\sqrt{3}}\left(3\right)\)

(1)(2)(3)\(=>A\le\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\)

\(A=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-3}}{z}\)

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.......

Đặt \(A=\frac{xy\sqrt{z-1}+xz\sqrt{y-2}+yz\sqrt{x-3}}{xyz}\)

\(\Rightarrow A=\frac{\sqrt{z-1}}{z}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{x-3}}{x}\)

\(\Rightarrow A=\frac{2.\sqrt{z-1}}{2z}+\frac{2.\sqrt{2}.\sqrt{y-2}}{2.\sqrt{2}.y}+\frac{2.\sqrt{3}.\sqrt{x-3}}{2.\sqrt{3}.x}\)\

\(\Rightarrow A\le\frac{z-1+1}{2z}+\frac{y-2+2}{2\sqrt{2}.y}+\frac{z-3+3}{2\sqrt{3}.x}\) ( ÁP DỤNG BĐT CÔ-SI )

\(\Rightarrow A\le\frac{z}{2z}+\frac{y}{2\sqrt{2}.y}+\frac{z}{2\sqrt{3}.z}\)

\(\Rightarrow A\le\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}=\frac{1}{2}+\frac{\sqrt{2}}{4}+\frac{\sqrt{3}}{6}\)

chuyển vế rồi thêm bớt cậu sẽ có rồi tìm được x=1 y=1 z=4

\(\left(x-2\sqrt{x}+1\right)+\left(y-2\sqrt{y}+1\right)+\left(z-4\sqrt{z}+4\right)=0\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-1\right)^2+\left(\sqrt{z}-2\right)^2=0\)

Đặt vế trái của BĐT cần chứng minh là P

Ta có:

\(P=\dfrac{\sqrt{xy+\left(x+y+z\right)z}+\sqrt{2\left(x^2+y^2\right)}}{1+\sqrt{xy}}=\dfrac{\sqrt{\left(x+z\right)\left(y+z\right)}+\sqrt{2\left(x^2+y^2\right)}}{1+\sqrt{xy}}\)

\(P\ge\dfrac{\sqrt{\left(\sqrt{xy}+z\right)^2}+\sqrt{\left(x+y\right)^2}}{1+\sqrt{xy}}=\dfrac{\sqrt{xy}+x+y+z}{1+\sqrt{xy}}=\dfrac{\sqrt{xy}+1}{1+\sqrt{xy}}=1\) (đpcm)

Dấu "=" xảy ra khi \(x=y\)

\(\dfrac{\sqrt{1+x^3+y^3}}{xy}>=\sqrt{\dfrac{3}{xy}}\)

\(\dfrac{\sqrt{1+y^3+z^3}}{yz}>=\sqrt{\dfrac{3}{yz}}\)

\(\dfrac{\sqrt{1+z^3+x^3}}{xz}>=\sqrt{\dfrac{3}{xz}}\)

=>\(VT>=\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)=3\sqrt{3}\)