Cho số dương $x, y$ thoả mãn : $xy\leq  1$ và $\frac{1}{x+1}+\frac{1}{y+1}\geq\frac{4}{3}   $
Tìm Min của $P=\frac{1}{x^2} +\frac{1}{y^2}-\frac{12}{x+y}  $
Với $xy \leq 1$ BĐT $\frac{1}{1+x}+\frac{1}{1+y}\leq \frac{2}{1+\sqrt{xy}}$ $\Leftrightarrow (1-\sqrt{xy})(\sqrt{x}-\sqrt{y})^{2} \geq 0$
Từ đó suy ra $\frac{2}{1+\sqrt{xy}} \geq \frac{4}{3}$ suy ra $\sqrt{xy}\leq \frac{1}{2}$
Ta có $P\geq \frac{2}{xy}-\frac{6}{\sqrt{xy}}$ Đặt $ t=\frac{1}{\sqrt{xy}}$ suy ra $t \geqslant 2$ suy ra $ P \geq f(t)$
Dễ thấy $f(t) $ đồng biến khi $t \geq  2$ suy $f(t)\geq f(2) = -4$ suy ra $Pmin = -4$ khi $x=y=\frac{1}{2}$
Trang hoctainha.vn trả lời nhanh- chi tiết. Rất cám ơn! –  vuvietphuong15 21-09-12 11:54 PM
Mình rất sợ những bài tìm Min- Max k hiểu sao sợ lắm ấy –  anh_chang_zaizai_90 21-09-12 11:40 PM

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