[tex]B=\frac{1}{x^3+y^3}+\frac{1}{xy}=\frac{1}{x^3+y^3}+\frac{3}{3xy(x+y)}\geq \frac{(1+\sqrt{3})^2}{x^3+y^3+3xy(x+y)}=\frac{(1+\sqrt{3})^2}{(x+y)^2}=(1+\sqrt{3})^2[/tex]
Dấu "=" xảy ra khi [tex]\left\{\begin{matrix} x+y=1\\ \frac{1}{x^3+y^3}=\frac{\sqrt{3}}{3xy} \end{matrix}\right.\Rightarrow \left\{\begin{matrix} x+y=1\\ \frac{1}{x^2-xy+y^2}=\frac{1}{\sqrt{3}xy} \end{matrix}\right.\Rightarrow \left\{\begin{matrix} x+y=1\\ x^2-(\sqrt{3}+1)xy+y^2=0 \end{matrix}\right.\Rightarrow \left\{\begin{matrix} x+y=1\\ x=\frac{\sqrt{3}+1+\sqrt{2\sqrt{3}}}{2}y \end{matrix}\right.\left\{\begin{matrix} (\frac{\sqrt{3}+1+\sqrt{2\sqrt{3}}}{2}+1)y=1\\ x=\frac{\sqrt{3}+1+\sqrt{2\sqrt{3}}}{2}y \end{matrix}\right.\Rightarrow \left\{\begin{matrix} y=\frac{2}{\sqrt{3}+1+\sqrt{2\sqrt{3}}+2}\\ x=\frac{\sqrt{3}+1+\sqrt{2\sqrt{3}}}{\sqrt{3}+1+\sqrt{2\sqrt{3}}+2} \end{matrix}\right.[/tex]
Vậy Min B = [TEX]4+2\sqrt{3}[/TEX]
(Phần tìm dấu "=" không cần cũng được nhé, bạn chỉ cần nói dấu "=" có xảy ra)