$4x^2+3x+5$
$= 4\left(x^2+\dfrac34x+\dfrac54\right)$
$= 4\left(x^2+\dfrac34x+\dfrac{9}{64}+\dfrac{71}{64}\right)$
$= 4\left[\left(x+\dfrac38\right)^2+\dfrac{71}{64}\right]$
$= 4\left(x+\dfrac38\right)^2+\dfrac{71}{16}$
Vì $4\left(x+\dfrac38\right)^2\ge 0\; \forall x\in \mathbb{R}$
$⇒ 4\left(x+\dfrac38\right)^2+\dfrac{71}{16}\ge \dfrac{71}{16}\; \forall x\in \mathbb{R}$
Vậy $\min = \dfrac{71}{16}$ khi $x+\dfrac38 = 0 ⇔ x=-\dfrac38$
