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Đặt \(\sqrt{x}=x;\sqrt{y}=y;\sqrt{z}=z\) cho dễ nhìn.
\(\Rightarrow\hept{\begin{cases}x+y+z=2\\x^2+y^2+z^2=2\end{cases}}\)
\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=4\)
\(\Leftrightarrow xy+yz+zx=1\)
Ta có:
\(x\left(1+y^2\right)\left(1+z^2\right)+y\left(1+z^2\right)\left(1+x^2\right)+z\left(1+x^2\right)\left(1+y^2\right)\)
\(=x^2y^2z+y^2z^2x+z^2x^2y+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y+x+y+z\)
\(=xyz\left(xy+yz+zx\right)+x^2\left(2-x\right)+y^2\left(2-y\right)+z^2\left(2-z\right)+2\)
\(=-2xyz+2\left(x^2+y^2+z^2\right)-\left(x^3+y^3+z^3-3xyz\right)+2\)
\(=-2xyz+6-\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(=-2xyz+6-2=-2xyz+4\)
Ta lại có:
\(\left(1+x^2\right)\left(1+y^2\right)\left(1+z^2\right)=x^2y^2z^2+x^2y^2+y^2z^2+z^2x^2+x^2+y^2+z^2+1\)
\(=x^2y^2z^2+\left(xy+yz+zx\right)^2-2xyz\left(xy+yz+zx\right)+3\)
\(=x^2y^2z^2-2xyz+4=\left(xyz-2\right)^2\)
\(\Rightarrow A=\sqrt{\left(xyz-2\right)^2}.\frac{4-2xyz}{\left(xyz-2\right)^2}\)
Tới đây bí :((
+ \(\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2=4\Rightarrow x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=4\)
\(\Rightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\)
+ \(x+1=x+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{z}\right)\)
+ Tương tự : \(y+1=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\); \(z+1=\left(\sqrt{x}+\sqrt{z}\right)\left(\sqrt{y}+\sqrt{z}\right)\)
+ \(P=\sqrt{\left(\sqrt{x}+\sqrt{y}\right)^2\left(\sqrt{y}+\sqrt{z}\right)^2\left(\sqrt{z}+\sqrt{x}\right)^2}\cdot\frac{\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)+\sqrt{y}\left(\sqrt{x}+\sqrt{z}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)}\)
\(=2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=2\)
\(\sqrt{x}+\sqrt{y}+\sqrt{z}=2\)
\(\Leftrightarrow x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{zx}=4\)
\(\Leftrightarrow2+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=4\)
\(\Leftrightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\)
Khi đó ta có : \(x+1=x+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
\(\Leftrightarrow x+1=\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)\)
\(\Leftrightarrow x+1=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\)
Tương tự : \(y+1=\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)\);
\(z+1=\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\)
Ta lần lượt xét các biểu thức :
+) \(\sqrt{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(=\sqrt{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}\)
\(=\sqrt{\left[\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\right]^2}\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\)
+) \(\frac{\sqrt{x}}{x+1}+\frac{\sqrt{y}}{y+1}+\frac{\sqrt{z}}{z+1}\)
\(=\frac{\sqrt{x}}{\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{x}+\sqrt{z}\right)}+\frac{\sqrt{y}}{\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}+\frac{\sqrt{z}}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)+\sqrt{y}\left(\sqrt{x}+\sqrt{z}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}\)
\(=\frac{2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\frac{2}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
Do đó ta có :
\(P=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\cdot\frac{2}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(P=2\)
Vậy...
1111111111111111111
\(VT=\Sigma\frac{xy+yz+zx}{xy}=3+\Sigma\frac{z\left(x+y\right)}{xy}\)
Đến đây để ý \(\frac{1}{2}\left[\frac{z\left(x+y\right)}{xy}+\frac{y\left(z+x\right)}{zx}\right]\ge\sqrt{\frac{\left(z+x\right)\left(x+y\right)}{x^2}}\left(\text{AM - GM}\right)\)
Là xong.
\(\frac{18\sqrt{2}}{3}=6\sqrt{2}\)
đặt mẫu số = Pain
áp dụng BDT cô si shaw ta có
\(\frac{1}{\sqrt{x\left(y+z\right)}}+\frac{1}{\sqrt{y\left(z+x\right)}}+\frac{1}{\sqrt{z\left(x+y\right)}}\ge\frac{9}{Pain}\)
áp dụng BDT cô si ta có ( thêm 2)
\(\sqrt{2x\left(y+z\right)}\le\frac{\left(2x+y+z\right)}{2}\)
\(\sqrt{2y\left(z+x\right)}\le\frac{\left(2y+z+x\right)}{2}\)
\(\sqrt{2z\left(x+y\right)}\le\frac{\left(2z+x+y\right)}{2}\)
+ lại và rút cái căn 2 ở VT và Tính VP ta được
\(\sqrt{2}\left(Pain\right)\le\frac{4}{2}\left(x+y+z\right)\) (x+y+z=18 căn 2)
\(\sqrt{2}\left(Pain\right)\le2\left(18.\sqrt{2}\right)\) ( rút gọn căn 2 với căn 2 )
\(Pain\le36\)
vì Pain năm ở dưới mẫu suy ra dấu \(\le\) thành dấu \(\ge\)
thay vào ta được
\(\frac{9}{Pain}\ge\frac{9}{36}=\frac{1}{4}\)