Đặt bài toán phụ : Chứng minh nếu \(a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\)

Thật vậy :

 \(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(a+b+c=0\Rightarrow\left(a+b+c\right)^3=0\)

\(a+b=-c\)

\(b+c=-a\)

\(c+a=-b\)

\(\Rightarrow\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(=-3\left(-c\right)\left(-b\right)\left(-a\right)\)

\(=3abc\)

Trở lại bài toán chính :

Ta có:

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Rightarrow\frac{yz}{xyz}+\frac{xz}{xyz}+\frac{xy}{xyz}=0\)

\(\Rightarrow\frac{yz+xz+xy}{xyz}=0\)

\(\Rightarrow xy+xz+yz=0\)

\(\Rightarrow\left(xy\right)^3+\left(xz\right)^3+\left(yz^3\right)=3\left(xy\right)\left(xz\right)\left(yz\right)=3x^2y^2z^2\)

Lại có:

\(P=\frac{xy.y^2x^2}{x^2y^2z^2}+\frac{xz.z^2.x^2}{x^2y^2z^2}+\frac{z^2.y^2.yz}{x^2y^2z^2}\)

\(=\frac{\left(xy\right)^3}{x^2y^2z^2}+\frac{\left(xz\right)^3}{x^2y^2z^2}+\frac{\left(yz\right)^3}{x^2y^2z^2}\)

\(=\frac{\left(xy\right)^3+\left(xz\right)^3+\left(yz^3\right)}{x^2y^2z^2}\)

Thay \(\left(xy\right)^3+\left(xz\right)^3+\left(yz^3\right)=3x^2y^2z^2;\)ta có:

\(P=\frac{3x^2y^2z^2}{x^2y^2z^2}\)

\(=3\)

Vậy \(P=3.\)