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\(\frac{x}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{z}\right)}+\frac{y}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{x}\right)}+\)\(\frac{z}{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{z}-\sqrt{y}\right)}\)
\(=-\frac{x}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{z}-\sqrt{x}\right)}-\frac{y}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)\(-\frac{z}{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)
\(=\frac{-x\left(\sqrt{y}-\sqrt{z}\right)-y\left(\sqrt{z}-\sqrt{x}\right)-z\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(=\frac{-x\sqrt{y}+x\sqrt{z}-y\sqrt{z}+y\sqrt{x}-z\sqrt{x}+z\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(=\frac{-\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)+\sqrt{z}\left(x-y\right)-z\left(\sqrt{x}-y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(=\frac{-\sqrt{xy}+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)-z}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(=\frac{-\sqrt{xy}+\sqrt{xz}+\sqrt{yz}-z}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(=\frac{\sqrt{y}\left(\sqrt{z}-\sqrt{x}\right)-\sqrt{z}\left(\sqrt{z}-\sqrt{x}\right)}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(=\frac{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{y}-\sqrt{z}\right)}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
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
\(\frac{9}{Pain}\ge\frac{9}{36}=\frac{1}{4}\)
Ta có:
\(1+x^2=xy+yz+zx+x^2=\left(x+y\right)\left(x+z\right)\)
\(1+y^2=xy+yz+xz+y^2=\left(y+z\right)\left(x+y\right)\)
\(1+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\)
Thay vào A được:
\(P=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}+y\sqrt{\frac{\left(x+z\right)\left(y+z\right)\left(x+y\right)\left(x+z\right)}{\left(y+z\right)\left(x+y\right)}}\)\(+z\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(x+z\right)\left(y+z\right)}}\)
\(=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)
\(=xy+xz+xy+yz+xz+zy\)
\(=2\left(xy+yz+xz\right)\)
\(=2\)(do xy+yz+xz=1)
=>Đpcm
Dạng toán này rất nhiều bạn hỏi rồi: thay \(xy+yz+zx=1\) vào các căn thức rồi phân tích đa thức thành nhân tử.
a) Ta có : \(1+x^2=xy+yz+zx+x^2=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(z+x\right)\)
b) \(\Sigma\left(x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\right)=\Sigma\left(x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right).\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\right)\)
\(=\Sigma\left(x\left(y+z\right)\right)=xy+xz+xy+yz+zx+zy=2\left(xy+yz+zx\right)=2\)
Côsi: \(\sqrt{x\left(y+z\right)}=\frac{1}{2\sqrt{2}}.2.\sqrt{2x}.\sqrt{y+z}\le\frac{1}{2\sqrt{2}}\left(2x+y+z\right)\)
\(\Rightarrow\frac{1}{\sqrt{x\left(y+z\right)}}\ge\frac{2\sqrt{2}}{2x+y+z}\)
Tương tự các cái kia.
\(\Rightarrow VT\ge2\sqrt{2}\left(\frac{1}{2x+y+z}+\frac{1}{2y+z+x}+\frac{1}{2z+x+y}\right)\)
\(\ge2\sqrt{2}.\frac{9}{2x+y+z+2y+z+x+2z+x+y}=\frac{18\sqrt{2}}{4\left(x+y+z\right)}=\frac{1}{4}\)
\(\sum\frac{1}{\sqrt{x\left(y+z\right)}}=\sum\frac{\sqrt{2}}{\sqrt{2x}.\sqrt{y+z}}\ge\sum\frac{2\sqrt{2}}{2x+y+z}\ge2\sqrt{2}.\frac{9}{\sum\left(2x+y+z\right)}=\frac{18\sqrt{2}}{4\left(x+y+z\right)}=\frac{1}{4}\)