Lời giải:
Ta có:
\(P=\frac{1}{x^2+y^2}+\frac{2}{xy}-4xy=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\left(\frac{1}{4xy}-4xy\right)+\frac{5}{4xy}\)
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)[(x^2+y^2)+2xy]\geq (1+1)^2\)
\(\Rightarrow \frac{1}{x^2+y^2}+\frac{1}{2xy}\geq \frac{4}{(x+y)^2}=4(1)\)
Áp dụng BĐT Cô-si: \(1=x+y\geq 2\sqrt{xy}\Rightarrow xy\leq \frac{1}{4}\)
\(\Rightarrow 1-4xy\geq 0\)
Do đó:
\(\frac{1}{4xy}-4xy=\frac{1-(16x^2y^2)}{4xy}=\frac{(1-4xy)(1+4xy)}{4xy}\geq 0(2)\)
\(xy\leq \frac{1}{4}\Rightarrow \frac{5}{4xy}\geq \frac{5}{4.\frac{1}{4}}=5(3)\)
Từ \((1);(2);(3)\Rightarrow P\geq 4+0+5=9\)
Vậy \(P_{\min}=9\)
Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)
