Đặt \(\left(a;b;c\right)=\left(x;y;\frac{1}{z}\right)\Rightarrow ab^2+bc^2+ca^2=3\)
\(P=\frac{1}{a^4+b^4+c^4}\)
Ta có:
\(a^4+b^4+b^4+1\ge4ab^2\)
\(b^4+c^4+c^4+1\ge4bc^2\)
\(c^4+a^4+a^4+1\ge4ca^2\)
\(\Rightarrow3\left(a^4+b^4+c^4\right)+3\ge4\left(ab^2+bc^2+ca^2\right)=12\)
\(\Rightarrow a^4+b^4+c^4\ge3\)
\(\Rightarrow P\le1\)