Question

Chứng minh rằng nếu /x/>=3;/y/>=3;/z/>=3

Thì A=\(\frac{x\cdot y+y\cdot z+z\cdot x}{x\cdot y\cdot z}\)<=1

Answer 1

\(A\le\left|A\right|=\dfrac{\left|xy+yz+xz\right|}{\left|xyz\right|}\)

Áp dụng: \(\left|a+b+c\right|\le\left|a\right|+\left|b\right|+\left|c\right|\)

\(\left|A\right|\le\dfrac{\left|xy\right|+\left|yz\right|+\left|xz\right|}{\left|xyz\right|}=\dfrac{1}{\left|x\right|}+\dfrac{1}{\left|y\right|}+\dfrac{1}{\left|z\right|}\)

\(\le\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=1\)

Ta có đpcm. Dấu "=" khi \(x=y=z=3\)

Answer 2

Thêm 1 hướng suy nghĩ khác
Ta có: \(\left|x\right|\ge3;\left|y\right|\ge3;\left|z\right|\ge3\)

\(\Rightarrow0< \dfrac{1}{\left|x\right|}\le\dfrac{1}{3};0< \dfrac{1}{\left|y\right|}\le\dfrac{1}{3};0< \dfrac{1}{\left|z\right|}\le\dfrac{1}{3}\)

Ta có:

\(A=\dfrac{xy+yz+zx}{xyz}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\le\dfrac{1}{\left|x\right|}+\dfrac{1}{\left|y\right|}+\dfrac{1}{\left|z\right|}\le\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=1\)