Ta có :
\(\frac{x^2+2x+3}{x^2+2}=\frac{2x^2+4-x^2+2x-1}{x^2+2}=\frac{2\left(x^2+2\right)-\left(x-1\right)^2}{x^2+2}=2-\frac{\left(x-1\right)^2}{x^2+2}\le2\)
\(\frac{x^2+2x+3}{x^2+2}=\frac{\frac{1}{2}x^2+1+\frac{1}{2}x^2+2x+2}{x^2+2}=\frac{\frac{1}{2}\left(x^2+2\right)+\frac{1}{2}\left(x+2\right)^2}{x^2+2}=\frac{1}{2}+\frac{2\left(x+2\right)^2}{x^2+2}\ge\frac{1}{2}\)