Fun with quadratics

We must have $a>0$ , because for negative a we will always get a negative value for large enough $x$ . Hence also $b>0$ .

If the function does not touch the x-axis we can decrease c until it does so, thereby decreasing the expression we want to minimize. So we can assume that $f(x)$ just touches the x-axis for some value of x, so that $b^2=4ac$ . We see that $c>0$ and we can set $b=2 \sqrt{ac}$

The value we want to minimize is $m=\frac{a+b+c}{b-a}=1+\frac{2a+c}{b-a}= 1 +\frac{2+\frac ca}{\frac ba-1}=1+\frac{2+\frac ca}{2\sqrt{\frac ca}-1}=1+\frac{2+r^2}{2r-1}$

seting $r=\sqrt{\frac ca}$ . By the quotient rule $\frac {dm}{dr} =0$ when $2r(2r-1)-2(2+r^2)=0$ which simplifies to $r^2-r-2=0$ with solutions $r=-1$ or $r=2$ . Since $r=\frac{b}{2a}$ , discard the negative value. For $r=2$ , we get $\boxed{m=3}$

If $f(x) \ge 0$ for all $x \in \mathbb{R}$ , two necessary & sufficient conditions that need to be invoked are:

$a > 0$ AND $b^2 - 4ac \le 0$ .

Let $a = b - k$ (where $k \in (0,b)$ ), and $c \ge \frac{b^2}{4a} = \frac{b^2}{4(b-k)}$ . Let the function $g(k)$ be:

$g(k) = \frac{(b-k) + b + \frac{b^2}{4(b-k)}}{b - (b-k)} = \frac{(3b - 2k)^2}{4k(b-k)}$ (i)

Differentiating (i) with respect to $k$ yields:

$g'(k) = \frac{(3b-2k)(12b^2 - 16bk)}{(4bk-4k^2)^2} = 0 \Rightarrow k = \frac{3b}{4}, \frac{3b}{2}$ (ii)

Since we require $0 < k < b$ , we only admit $k = \frac{3b}{4}$ . A second derivative check at this critical value gives:

$g''(\frac{3b}{4}) = \frac{128}{3b^2} > 0$

hence, a global minimum for $k \in (0,b)$ . Our final minimum value computes to:

$g(\frac{3b}{4}) = \frac{(3b - \frac{3b}{2})^2}{4(\frac{3b}{4})(b - \frac{3b}{4})} = \frac{9}{4} \cdot \frac{4}{3} = \boxed{3}.$