## Calculus (3rd Edition)

$$x^{2}+3 x+3\geq 0,\ \ \text{for all } x$$
We can rewrite the expression as: \begin{align*} x^{2}+3 x+3&=\left(x+\frac{3}{2}\right)^2-\frac{9}{4} +3\\ &= \left(x-\frac{3}{2}\right)^{2}+\frac{3}{4} \end{align*} We see that $\left(x-\frac{3}{2}\right)^{2} \geq 0$ for all $x$, so $$x^{2}+3 x+3\geq 0,\ \ \text{for all } x$$