Câu 3:
Các BĐT cần CM: $\sqrt{2(x^{2}+y^{2})}\geqslant x+y$
$\frac{7}{2}(x^{2}+y^{2})+xy\geq 2(x+y)^{2}$
Ta có: $P=\frac{xy}{x^{2}+y^{2}}+(\frac{1}{x}+\frac{1}{y})\sqrt{2(x^{2}+y^{2})}\geq \frac{2(x+y)^{2}}{x^{2}+y^{2}}+\frac{(x+y)^{2}}{xy}=\frac{2(x+y)^{2}}{x^{2}+y^{2}}+\frac{2(x+y)^{2}}{2xy}-\frac{7}{2}\geq 2(\frac{4(x+y)^{2}}{(x+y)^{2}})-\frac{7}{2}=8-\frac{7}{2}=\frac{9}{2}$
Dấu {"="} xảy ra $\Leftrightarrow x=y$