Điều kiện: `x,y>=0`
Ta có:
`A=`$x\sqrt[]{x}+y\sqrt[]{y}$
$=\sqrt[]{x^3}+\sqrt[]{y^3}$
$=(\sqrt[]{x}+\sqrt[]{y})(x-\sqrt[]{xy}+y)$
$=x-\sqrt[]{xy}+y$
Vì `x; y>=0=>A>=0`
`<=>`$x-\sqrt[]{xy}+y\ge0$
`<=>`$x+2\sqrt[]{xy}+y-3\sqrt[]{xy}\ge0$
`<=>`$(\sqrt[]{x}+\sqrt[]{y})^2-3\sqrt[]{xy}\ge0$
`<=>`$1-3\sqrt[]{xy}\ge0$
Vì $-3\sqrt[]{xy}\le0$
`=>`$1-3\sqrt[]{xy}\le1$
Dấu `=` xảy ra `<=>xy=0<=>`\(\left[ \begin{array}{l}x=0;y=1\\y=0; x=1\end{array} \right.\)
Vậy ...
