Chứng minh rằng nếu IxI≥3,IyI≥3,IzI≥3 thì A=(y+yz+zx)/(xyz)
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\(\left\{{}\begin{matrix}\left|x\right|\ge3\\\left|y\right|\ge3\\\left|z\right|\ge3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|\dfrac{1}{x}\right|\le\dfrac{1}{3}\\\left|\dfrac{1}{y}\right|\le\dfrac{1}{3}\\\left|\dfrac{1}{z}\right|\le\dfrac{1}{3}\end{matrix}\right.\)
\(\left|A\right|=\left|\dfrac{xy+yz+xz}{xyz}\right|=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\le\left|\dfrac{1}{x}\right|+\left|\dfrac{1}{y}\right|+\left|\dfrac{1}{z}\right|\le\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=1\)
\(\Rightarrow A\le\left|A\right|\le1\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=3\)
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a. Ta có :
\(\left|x+y\right|\le\left|x\right|+\left|y\right|\Leftrightarrow\left(\left|x\right|+\left|y\right|\right)^2\ge\left|x+y\right|^2=\left(x+y\right)^2\)
\(\Leftrightarrow x^2+y^2+2\left|xy\right|\ge x^2+2xy+y^2\)
\(\Leftrightarrow2\left|xy\right|\ge2xy\Leftrightarrow\left|xy\right|\ge xy\) ( luôn đúng )
Dấu "=" xảy ra <=> x và y cùng dấu
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\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)
\(\Leftrightarrow\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}=\frac{x^2-y^2+xz-yz}{x-xyz-y+xyz}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-y}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\)
\(\Rightarrow\frac{x^2-yz}{x-xyz}=x+y+z\)
\(\Rightarrow x^2-yz=\left(x-xyz\right)\left(x+y+z\right)\)
\(\Rightarrow x^2-yz=x\left(x-xyz\right)+y\left(x-xyz\right)+z\left(x-xyz\right)\)
\(\Rightarrow x^2-yz=x^2-x^2yz+xy-xy^2z+xz-xyz^2\)
\(\Rightarrow-yz-xy-xz=-x^2yz-xy^2z-xyz^2\)
\(\Rightarrow-\left(yz+xy+xz\right)=-\left(x^2yz+xy^2z+xyz^2\right)\)
\(\Rightarrow yz+xy+xz=x^2yz+xy^2z+xyz^2\)
\(\Rightarrow yz+xy+xz=xyz\left(x+y+z\right)\)
Vậy nếu \(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\) thì \(yz+xy+xz=xyz\left(x+y+z\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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