Is this easy?

The roots of the above quadratic equation equal:

$x = \frac{-b \pm \sqrt{b^2 - 4a}}{2a}$ (i)

In order to have no distinct real roots, we require the discriminant in (i) to be $b^2 - 4a \leq 0$ . This ensures we have either one real repeated root OR a complex-conjugate pair of roots. Solving the above inequality for a in terms of b gives $a \geq \frac{b^2}{4}$ , and substituting this expression into $a - b$ results in the quadratic function:

$f(b) = a - b = \frac{b^2}{4} - b = \frac{1}{4}(b - 2)^2 - 1$ (ii)

The concave-up parabola described in (ii) has a global minimum at the vertex $(2,-1)$ , so $\boxed{-1}$ is the smallest value.

If two roots are not distinct real roots, they might have repeated roots.
$This~ means~ \Delta~ can ~be ~0. ~~~So ~\Delta=(b^2-4a)=0.~\implies~b^2=4a.~~Smallest~value~can~easily~be~seen~as~a=1~an~b=2. \\ So~ a-b=1-2= \Large~~~\color{#D61F06}{-1}$
.

Let $x_1$ and $x_2$ be the solutions of $ax^2 + bx + 1 = 0\;.$ Since they are not distinct real roots, the discriminant is non-positive, i.e. $\Delta = b^2 - 4a \le 0\;,$ and we can write $x_1$ and $x_2$ as follows: $x_1 = f+ig\;, \qquad x_2 = f-ig\;,$ with $f$ and $g$ real. Using Vieta's formulae, we can establish that $x_1 + x_2 = 2f = -\frac{b}{a}\;, \qquad x_1x_2 = f^2 + g^2 = \frac{1}{a}\;.$ and from these and the non-positivity of the discriminant we can write: $a - b = \frac{2f + 1}{f^2 + g^2}\;, \qquad \frac{f^2}{f^2 + g^2} \le 1\;.$ The second condition is automatically satisfied for every $f$ and $g$ . We can write $a - b = \frac{2f + 1}{f^2 + g^2} = \frac{(f + 1)^2}{f^2 + g^2} - \frac{f^2}{f^2 + g^2}\;,$ the difference of two positive number. The minimum can be obtained by making the first term to vanish, i.e. $f = -1$ , and making $g^2$ as small as possible, i.e. $g = 0$ . Therefore: $\boxed{a - b \ge -1}$