Explain why $3x^2+15x+20$ cannot be smaller than 1.

1. Complete the square for the given quadratic expression.

$3x^2 + 15x + 20$
$= 3\left(x^2 + 5x\right) + 20$
$= 3\left[\left(x + \frac{5}{2}\right)^2 - \left(\frac{5}{2}\right)^2\right] + 20$
$= 3\left(x + \frac{5}{2}\right)^2 - 3\left(\frac{25}{4}\right) + 20$
$= 3\left(x + \frac{5}{2}\right)^2 - \frac{75}{4} + \frac{80}{4}$
$= 3\left(x + \frac{5}{2}\right)^2 + \frac{5}{4}$

2. Use the completed square form to deduce the minimum value and conclude.

Since $\left(x + \frac{5}{2}\right)^2 \geqslant 0$ for all real values of $x$,
$3\left(x + \frac{5}{2}\right)^2 + \frac{5}{4} \geqslant \frac{5}{4}$
The minimum value of the expression is $\frac{5}{4}$ (or $1.25$).
Since $1.25 > 1$, the expression $3x^2 + 15x + 20$ cannot be smaller than 1.