nâng cao9, khó lắm?

H

hthtb22

$$\text{Ta có:}\\P=\frac{2}{xy}+\frac{3}{x^2+y^2}=\frac{1}{2xy}+3(\frac{1}{2xy}+\frac{1}{x^2+y^2})\\\text{Thấy}xy \le \frac{(x+y)^2}{4}=\frac{1}{4} \text{nên}\frac{1}{2xy} \ge 2\\ 3(\frac{1}{2xy}+\frac{1}{x^2+y^2}) \ge 3.\frac{9}{(x+y)^2}=27\text{(AM-GM cộng mẫu)}...$$
 
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