Pt⇔x2+2x+1+y2+4y+6−2√(x2+2x+1)(y2+4y+6)=2x2⇔(√x2+2x+1−√y2+4y+6)2=2x2
Ta có :√x2+2x+1=√(x+1)2⩾
\sqrt{y^{2}+4y+6}=\sqrt{(y+2)^2+2}\geqslant \sqrt{2}
\Rightarrow (\sqrt{x^{2}+2x+1}-\sqrt{y^{2}+4y+6})^2\geqslant 2
Dấu "="xảy ra\Leftrightarrow x+1=0;y+2=0 và x^2=1\Leftrightarrow \begin{cases}x=-1 \\ y=-2 \end{cases}
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