Let \(\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)=\left(a;b;c\right)\) we need prove:
\(\left\{{}\begin{matrix}a+b+c=1\\a^4+b^4+c^4\ge abc\\a,b,c\ne0\end{matrix}\right.\)
By AM-GM we have: \(\left\{{}\begin{matrix}a^4+b^4\ge2\sqrt{a^4b^4}=2a^2b^2\\b^4+c^4\ge2\sqrt{b^4c^4}=2b^2c^2\\c^4+a^4\ge2\sqrt{c^4a^4}=2c^2a^2\end{matrix}\right.\)
\(\Rightarrow a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2\left(1\right)\)
By AM-GM we have:
\(\left\{{}\begin{matrix}a^2b^2+b^2c^2=b^2\left(a^2+c^2\right)\ge b^2\cdot2\sqrt{a^2c^2}=2b^2ac\\b^2c^2+c^2a^2=c^2\left(b^2+a^2\right)\ge c^2\cdot2\sqrt{b^2a^2}=2c^2ab\\c^2a^2+a^2b^2=a^2\left(b^2+c^2\right)\ge a^2\cdot2\sqrt{b^2c^2}=2a^2bc\end{matrix}\right.\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2\ge b^2ac+c^2ab+a^2bc\)
\(=abc\left(a+b+c\right)=abc\left(a+b+c=1\right)\left(2\right)\)
From \((1);(2)\) we are done !!
