To prove that $a(a+b)+b(b+1)\geq -\frac {1}{3}$ is same as proving that $a(a+b)+b(b+1)+\frac {1}{3}\geq 0$.
Consider, $a(a+b)+b(b+1)+\frac {1}{3}$
Or, $b^2+b(a+1)+a^2+\frac {1}{3}$
The above can be considered as quadratic expression in b, whose coefficient of $b^2$ is 1.
Discriminant $\Delta$=$(a+1)^2-4.\left( {{a^2} + \frac{1}{3}} \right)$
= $-3a^2+2a-\frac {1}{3}$
= $ - 3\left( {{a^2} - \frac{2}{3}a + \frac{1}{9}} \right)$
= $ - 3{\left( {a - \frac{1}{3}} \right)^2}$
$ \le 0$
Since coefficient of $b^2$ is 1 and discriminant is $ \le 0$ for the quadratic expression in b, the quadratic expression itself is $\ge 0$.
