Ta có

$2011 . x^{2012} + 1 = \underbrace{x^{2012} + \cdots + x^{2012}}_{2011 \, \text{số hạng}} + 1$

$\geq 2012 \sqrt[2012]{\underbrace{x^{2012} . \dots . x^{2012}}_{2011 \, \text{số hạng} . 1}}$

$= 2012 \sqrt[2012]{x^{2012 . 2011}} = 2012 x^{2011}$

Làm tương tự ta có

$2011 y^{2012} + 1 \geq 2012 y^{2011}$

Suy ra

$2011 . x^{2012} + 1 + 2011 y^{2012} + 1 \geq 2012 x^{2011} + 2012 y^{2011}$

$<-> 2011(x^{2012} + y^{2012}) + 2 \geq 2012(x^{2011} + y^{2011})$

$<-> 2011(x^{2012} + y^{2012}) \geq 2011(x^{2011} + y^{2011}) + (x^{2011} + x^{2011}) -2$

Lại có

$x^{2011} + 2010 = x^{2011} + \underbrace{1 +\cdots + 1}_{2010 \, \text{số hạng}}$

$\geq  2011 \sqrt[2011]{x^{2011} . \underbrace{1 .\dots . 1}_{2010 \, \text{số hạng}}}$

$= 2011 \sqrt[2011]{x^{2011}} = 2011x$

CMTT ta có

$y^{2011} + 2010 \geq 2011 y$