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*) Tìm Min
Ta có:
$S=16x^2y^2+12(x^3+y^3)+34xy$
$=16x^2y^2+12(x+y)^3-36xy+34xy$
$=16x^2y^2-2xy+12$
$=16(xy-\frac{1}{16})^2+\frac{191}{16}\ge\frac{191}{16}$
Min$S=\frac{191}{16}\Leftrightarrow\left\{ \begin{array}{l} x+y=1\\ xy=\frac{1}{16} \end{array} \right. \Leftrightarrow (x,y)\in\{(\frac{2+\sqrt3}{4};\frac{2-\sqrt3}{4});(\frac{2-\sqrt3}{4};\frac{2+\sqrt3}{4})\}$
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