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Xét hiệu $x^8+y^8-x^6y^2-y^6x^2$=$(x^6-y^6)(x^2-y^2)$

=$(x^3+y^3)(x^3-y^3)(x+y)(x-y)$

=$(x+y)^2(x-y)^2(x^2+xy+y^2)(x^2-xy+y^2)$\geq0

\Rightarrow$x^8+y^8$\geq$x^2y^2(x^4+y^4)$

\Leftrightarrow$\dfrac{x^8+y^8}{x^2y^2}$\geq$x^4+y^4$

\Leftrightarrow$\dfrac{x^6}{y^2}+\dfrac{y^6}{x^2}$\geq$x^4+y^4$
 
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