Chủ đề này có 1 trả lời

[hanguyen225]

Tìm GTNN của M=$\frac{2}{xy}+\frac{3}{x^{2}+y^{2}}$

[DOTOANNANG]

$\frac{2}{xy}+ \frac{3}{x^{\,2}+ y^{\,2}}=$ $\frac{1}{2\,xy}+ \frac{1}{2\,xy}+ \frac{1}{2\,xy}+ \frac{1}{2\,xy}$ $+ \frac{1}{x^{\,2}+ y^{\,2}}+ \frac{1}{x^{\,2}+ y^{\,2}}+ \frac{1}{x^{\,2}+ y^{\,2}}$ $\geqq \frac{1}{xy+ \frac{x^{\,2}}{2}+ \frac{y^{\,2}}{2}}+ \frac{1}{2\,xy}+ \frac{1}{2\,xy}+ \frac{1}{2\,xy}+$ $+ \frac{1}{x^{\,2}+ y^{\,2}}+ \frac{1}{x^{\,2}+ y^{\,2}}+ \frac{1}{x^{\,2}+ y^{\,2}}\geqq$ $\frac{\left ( 1+ 1+ 1+ 1+ 1+ 1+ 1 \right )^{\,2}}{xy+ \frac{x^{\,2}}{2}+ \frac{y^{\,2}}{2}+ 2\,xy+ 2\,xy+ 2\,xy+ x^{\,2}+ y^{\,2}+ x^{\,2}+ y^{\,2}+ x^{\,2}+ y^{\,2}}\geqq \frac{\left ( 2.\,2+ 3 \right )^{\,2}}{\frac{2.\,2+ 3}{2}\,\left ( x+ y \right )^{\,2}}$

$\lceil$ Bổ đề Titu (!) $\rfloor$

 

$\lfloor$ Tổng quát (!) $\rceil$ Với $2\,m\geqq n$ thì: $\frac{m}{xy}+ \frac{n}{x^{\,2}+ y^{\,2}}\geqq \frac{m+ \frac{n}{2}}{\left ( x+ y \right )^{\,2}}$ hay: $\frac{\left ( x- y \right )^{\,2} \left [ m\,\left ( x- y \right )^{\,2}+ \left (2\,m- n \right )xy \right ]}{xy\,\left (x+ y \right )^{\,2}\left ( x^{\,2}+ y^{\,2} \right )}\geqq 0$