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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 \(\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-2}=\sqrt{2}\\\sqrt{z-3}=\sqrt{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=4\\z=6\end{matrix}\right.\)
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\(T=21x+\frac{21}{y}+3y+\frac{3}{x}\)
\(=\frac{x}{3}+\frac{3}{x}+\frac{21}{y}+\frac{7y}{3}+\frac{62x}{3}+\frac{2y}{3}\)
\(\ge2\sqrt{\frac{x}{3}\cdot\frac{3}{x}}+2\sqrt{\frac{21}{y}\cdot\frac{7y}{3}}+\frac{62\cdot3}{3}+\frac{2\cdot3}{3}\)
\(=2+14+62+2=80\)
\(T=80\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{3}=\frac{3}{x}\\\frac{21}{y}=\frac{7y}{3}\\x=3\\y=3\end{matrix}\right.\Leftrightarrow x=y=3\)
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a)\(M=\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}.\left(\sqrt{x}+1\right)\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}-2}\)
b)\(\frac{1}{M}=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)
Ta có: \(\sqrt{x}\ge0,\forall x\ge0\)
\(\Leftrightarrow\sqrt{x}+1\ge1\)
\(\Leftrightarrow\frac{1}{\sqrt{x}+1}\le1\)
\(\Leftrightarrow\frac{3}{\sqrt{x}+1}\le3\)
\(\Leftrightarrow-\frac{3}{\sqrt{x}+1}\ge-3\)
\(\Leftrightarrow1-\frac{3}{\sqrt{x}+1}\ge-2\)
Dấu "=" xảy ra khi x=0
Vậy \(Min_{\frac{1}{M}}=-2\) khi x=0
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\(A=\frac{1}{x^3+y^3}+\frac{1}{xy}=\frac{1}{\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]}+\frac{1}{xy}=\frac{1}{1-3xy}+\frac{1}{xy}\)
Theo BĐT Cô si ta có :
\(A=\frac{1}{1-3xy}+\frac{3}{3xy}\ge\frac{\left(1+\sqrt{3}\right)^2}{1-3xy+3xy}=4+2\sqrt{2}\)
Vậy BĐT đã được chứng minh .
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\(Q=\Sigma\frac{x^4}{x^2+\sqrt{xy.zx}}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+xy+yz+zx}\ge\frac{x^2+y^2+z^2}{2}\ge\frac{\left(x+y+z\right)^2}{6}=\frac{3}{2}\)
Dấu "=" xảy ra khi x=y=z=1
Áp dụng bđt Cauchy cho 2 số dương:
\(A=x+\frac{1}{x}=\frac{8x}{9}+\frac{x}{9}+\frac{1}{x}\ge\frac{8.3}{9}+2\sqrt{\frac{x}{9}.\frac{1}{x}}=\frac{10}{3}\)
Dấu "=" xảy ra khi x=3
\(A=\left(\frac{x}{9}+\frac{1}{x}\right)+\frac{8}{9}x\)
\(\ge2\sqrt{\frac{x}{9}.\frac{1}{x}}+\frac{8}{9}\times3\) \(=2\times\frac{1}{3}+\frac{8}{3}=\frac{10}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{x}{9}=\frac{1}{x}\Leftrightarrow x=3\left(tmđk\right)\)