Ta có :[tex]5x^{2}+2y^{2}+4xy-2x+4y+2015[/tex]
=[tex](4x^{2}+4xy+y^{2})+(x^{2}-2x+1)+(y^{2}+4y+4)+2010[/tex]
=[tex](2x+y)^{2}+(x-1)^{2}+(y+2)^{2}+2010[/tex]
Vì [tex](2x+y)^{2}\geq 0[/tex] ; [tex](x-1)^{2}\geq 0[/tex] ; [tex](2+y)^{2}\geq 0[/tex]
=>[tex](2x+y)^{2}+(x-1)^{2}+(y+2)^{2}\geq 0[/tex]
=>[tex](2x+y)^{2}+(x-1)^{2}+(y+2)^{2}+2010\geq 2010[/tex]
Hay [tex]5x^{2}+2y^{2}+4xy-2x+4y+2015\geq 2010[/tex]
Dấu "=" xảy ra khi : x = 1 ; y = -2
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