Question

Tìm nghiệm nguyên dương

xyz(\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\)) = 3

Answer 1

\(xyz\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{x^2}\right)=3\)

\(\Leftrightarrow\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}=3\)

Áp dụng bđt AM-GM có:

\(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\ge3\sqrt{\dfrac{yz}{x}.\dfrac{xz}{y}.\dfrac{xy}{z}}\)\(\Leftrightarrow3\ge3\sqrt{xyz}\Leftrightarrow1\ge xyz\)

Mà \(x,y,z\in N\)*\(\Rightarrow xyz\in N\)*\(\Rightarrow xyz=1\)

\(\Rightarrow x\inƯ\left(1\right);y\inƯ\left(1\right);z\inƯ\left(1\right)\)

\(\Rightarrow x=y=z=1\)(Tm)

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