Ta có : \(x^2+xy+y^2+1>0\)
\(\Leftrightarrow x^2+xy+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2+1>0\)
\(\Leftrightarrow\left(x^2+xy+\dfrac{1}{4}y^2\right)+\left(\dfrac{3}{4}y^2+1\right)>0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}y\right)^2+\left(\dfrac{3}{4}y^2+1\right)>0\)
Do : \(\left(x+\dfrac{1}{2}y\right)^2\ge0\) \(\forall xy\) và \(\dfrac{3}{4}y^2+1\ge1>0\) \(\forall y\)
\(\Rightarrow\left(x+\dfrac{1}{2}y\right)^2+\left(\dfrac{3}{4}y^2+1\right)>0\)
Vậy : \(x^2+xy+y^2+1>0\) ( đpcm )
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