x^2 + 3x + 4 = x^2 + 2.(3/2).x + 9/4 - 9/4 + 4
= (x+3/2)^2 + 7/4 >= 7/4
<=> A >= 49/16.
Dấu "=" xảy ra <=> x = -3/2
\(x^2+3x+4=\left(x+\frac{3}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\left(x^2+3x+4\right)^2\ge\left(\frac{7}{4}\right)^2=\frac{48}{16}\)
Dấu '=' xảy ra khi :
\(x+\left(\frac{3}{2}\right)=0\Leftrightarrow x=-\frac{3}{2}\)