[TEX]1. (x+y+1)xy = x^2 + y^2[/TEX]
[TEX]\Leftrightarrow \frac{1}{x} + \frac{1}{y} = \frac{1}{x^2} - \frac{1}{xy} + \frac{1}{y^2} = \bigg(\frac{1}{x} - \frac{1}{y} \bigg)^2 + \frac{3}{4y^2} >0[/TEX]
[TEX]\Rightarrow \frac{1}{x} + \frac{1}{y} > 0[/TEX]
[TEX]\Leftrightarrow \frac{1}{x} + \frac{1}{y} = \frac{1}{x^2} - \frac{1}{xy} + \frac{1}{y^2} = \bigg( \frac{1}{x} + \frac{1}{y} \bigg)^2 - \frac{3}{xy} \leq \bigg(\frac{1}{x} + \frac{1}{y}\bigg)^2 - \frac{3}{4}\bigg( \frac{1}{x} + \frac{1}{y} \bigg)^2[/TEX]
[TEX]\Leftrightarrow \frac{1}{x} + \frac{1}{y} \leq \frac{1}{4} \bigg(\frac{1}{x} + \frac{1}{y} \bigg)^2[/TEX]
[TEX]\Leftrightarrow 4\bigg(\frac{1}{x} + \frac{1}{y} \bigg) - \bigg(\frac{1}{x} + \frac{1}{y}\bigg)^2 \leq 0 [/TEX]
[TEX]\Leftrightarrow \bigg(\frac{1}{x} + \frac{1}{y} \bigg)\bigg( 4 - \bigg(\frac{1}{x} + \frac{1}{y}\bigg) \bigg) \leq 0[/TEX]
[TEX]\frac{1}{x} + \frac{1}{y} \leq 4[/TEX]
[TEX]A = \frac{1}{x^3} + \frac{1}{y^3} = \bigg(\frac{1}{x} + \frac{1}{y}\bigg)\bigg(\frac{1}{x^2} - \frac{1}{xy} + \frac{1}{y^2}\bigg) = \bigg(\frac{1}{x} + \frac{1}{y} \bigg)^2 \leq 16[/TEX]