[tex]B=x^3+y^3+8(x^4+y^4)+\frac{2}{xy}\geq (x^3+y^3+\frac{1}{8})-\frac{1}{8}+8\frac{(x^2+y^2)^2}{2}+\frac{2}{xy}\geq \frac{3}{2}xy+\frac{2}{xy}-\frac{1}{8}+8\frac{[\frac{(x+y)^2}{2}]^2}{2}=(\frac{3}{2}xy+\frac{3}{32xy})-\frac{1}{8}+(x+y)^4+\frac{61}{32xy}\geq \frac{3}{16}-\frac{1}{8}+1+\frac{61}{32\frac{(x+y)^2}{4}}=\frac{37}{4}[/tex]