Answer
$$ x^{2}+3 x+3\geq 0,\ \ \text{for all } x$$
Work Step by Step
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 $$ x^{2}+3 x+3\geq 0,\ \ \text{for all } x$$
