Xét biểu thức \(\frac{14x^2-8x+9}{3x^2+6x+9}=\frac{14x^2-8x+9}{3x^2+6x+9}-\frac{2}{3}+\frac{2}{3}=\frac{3\left(14x^2-8x+9\right)-2\left(3x^2+6x+9\right)}{3\left(3x^2+6x+9\right)}+\frac{2}{3}=\frac{36x^2-36x+9}{3\left(3x^2+6x+9\right)}+\frac{2}{3}=\frac{\left(6x-3\right)^2}{3\left(3x^2+6x+9\right)}+\frac{2}{3}\ge\frac{2}{3}\)Đẳng thức xảy ra khi x = 1/2