Bài này mình làm tắt nha:
A=$\frac{1}{x^{2}+y^{2}}+\frac{1}{xy}+xy$
=$(\frac{1}{x^{2}+y^{2}}+\frac{1}{2xy})+\frac{1}{2xy}+xy\geq \frac{2}{\sqrt{(x^{2}+y^{2})2xy}}+\frac{1}{2xy}+8xy-7xy $ (theo bdt Cauchy)
$\geq $ $\frac{4}{(x+y)^{2}}+2\sqrt{\frac{1}{2xy}.8xy}-7xy\geq 4+4-7xy (do x+y$\leq $1)
ta có: $1\geq x+y\geq 2\sqrt{xy}\Leftrightarrow 1\geq 4xy \Leftrightarrow -7xy\geq -1,75$
$\Rightarrow A\geq 4+4-1,75=6,25$
Vậy $A_{min}=6,25$