Có : \(2x^2+3x+2\)
\(\Leftrightarrow\) \(\left(x^2+2x+1^2\right)+\left(x^2+x+1^2\right)\)
\(\Leftrightarrow\) \(\left(x^2+2.x.1+1^2\right)\) + \(\left(x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\right)\)
\(\Leftrightarrow\) \(\left(x+1\right)^2+\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+1\right)^2\ge0và\left(x+\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow\) \(\left(x+1\right)^2+\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
Vậy \(2x^2+3x+2>0\left(\forall_x\right)\)