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Ta có: \(3\sqrt{x+2y-1}=\sqrt{9\left(x+2y-1\right)}\le\frac{9+x+2y-1}{2}\)
\(=\frac{x+2y}{2}+4\Leftrightarrow3\sqrt{x+2y-1}-4\le\frac{x+2y}{2}\)(1)
Tương tự ta có: \(3\sqrt{y+2z-1}\le\frac{y+2z}{2}\left(2\right);3\sqrt{z+2x-1}\le\frac{z+2x}{2}\left(3\right)\)
Cộng theo vế của 3 BĐT (1), (2), (3), ta được:
\(T=\frac{x}{3\sqrt{x+2y-1}-4}+\frac{y}{3\sqrt{y+2z-1}-4}+\frac{z}{3\sqrt{z+2x-1}-4}\)
\(\ge\frac{2x}{x+2y}+\frac{2y}{y+2z}+\frac{2z}{z+2x}\)\(=2\left(\frac{x^2}{x^2+2xy}+\frac{y^2}{y^2+2yz}+\frac{z^2}{z^2+2zx}\right)\)
\(\ge2.\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}=2.\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=2\)(Theo BĐT Bunhiacopxki dạng phân thức)
Đẳng thức xảy ra khi \(x=y=z=\frac{10}{3}\)
\(Q=\Sigma\frac{x^4}{x^2+\sqrt{xy.zx}}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+xy+yz+zx}\ge\frac{x^2+y^2+z^2}{2}\ge\frac{\left(x+y+z\right)^2}{6}=\frac{3}{2}\)
Dấu "=" xảy ra khi x=y=z=1
Ta có : \(\left\{{}\begin{matrix}x\ge1\\y\ge2\\z\ge3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\sqrt{x-1}\ge0\\\sqrt{y-2}\ge0\\\sqrt{z-3}\ge0\end{matrix}\right.\Rightarrow\sqrt{x-1}+\sqrt{y-2}+\sqrt{z-3}\ge0\)
Đặt \(\sqrt{x-1}=a;\sqrt{y-2}=b;\sqrt{z-3}=c\)
\(\Rightarrow A=\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\)
\(\sum\frac{a}{a^2+1}=\sum\left(a-\frac{a^3}{a^2+1}\right)\ge\sum\left(a-\frac{a}{2}\right)=\frac{a+b+c}{2}\)
\(\Rightarrow A\ge\frac{\sqrt{x-1}+\sqrt{y-2}+\sqrt{z-3}}{2}=0\)
Vậy \(MIN_A=0\) khi \(x=1;y=2;z=3\)
\(A=\frac{1.\sqrt{x-1}}{x}+\frac{1}{\sqrt{2}}.\frac{\sqrt{2}.\sqrt{y-2}}{y}+\frac{1}{\sqrt{3}}.\frac{\sqrt{3}.\sqrt{z-3}}{z}\)
\(A\ge\frac{1+x-1}{2x}+\frac{1}{\sqrt{2}}\left(\frac{2+y-2}{2y}\right)+\frac{1}{\sqrt{3}}\left(\frac{3+z-3}{2z}\right)=\frac{6+3\sqrt{2}+2\sqrt{3}}{12}\)
\(\Rightarrow A_{min}=\frac{6+3\sqrt{2}+2\sqrt{3}}{12}\) khi \(\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-2}=\sqrt{2}\\\sqrt{z-3}=\sqrt{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=4\\z=6\end{matrix}\right.\)