Ta có:
\(VT=xyz\left(\dfrac{1}{z^2}+\dfrac{1}{y^2}+\dfrac{1}{x^2}\right)=3\)
\(\Rightarrow xyz>0\)
Áp dụng BĐT Cauchy cho 3 số dương ta có:
\(3=\dfrac{xyz}{z^2}+\dfrac{xyz}{y^2}+\dfrac{xyz}{z^2}\ge3\sqrt[3]{xyz}\)
\(\Rightarrow1\ge\sqrt[3]{xyz}\)
\(\Rightarrow1\ge xyz\)
\(\Rightarrow0< xyz\le1\)
Vì x, y, z nguyên nên \(xyz=1\)
Mà \(xyz=1\) \(\Rightarrow\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\ge3\sqrt[3]{xyz}=3\)
Dấu "=" xảy ra khi \(x^2+y^2+z^2\)
\(\Leftrightarrow\left|x\right|=\left|y\right|=\left|z\right|\)
Từ đó tìm ra 4 nghiệm là \(\left(x,y,z\right)=\left(1,1,1\right);\left(1,-1,-1\right);\left(-1,1,-1\right);\left(-1,-1,1\right).\)
