$\quad x^5 + y^5 - x^4y - xy^4$
$= (x^5 - x^4y) + (y^5 - xy^4)$
$= x^4(x-y) + y^4(y-x)$
$= (x-y)(x^4 - y^4)$
$= (x-y)(x^2 - y^2)(x^2+ y^2)$
$= (x-y)^2(x+y)(x^2 + y^2)$
Ta có:
$\begin{cases}(x-y)^2\geqslant 0\quad \forall x,y\\ x+y\geqslant 0\quad (gt)\\x^2 + y^2 \geqslant 0\quad \forall x,y\end{cases}$
Do đó: $(x-y)^2(x+y)(x^2 + y^2)\geqslant 0$
Hay $x^5 + y^5 - x^4y - xy^4 \geqslant 0$
Dấu $=$ xảy ra $\Leftrightarrow x = y = 0$
