3A. Given the set of quadratic functions $f(x)=a x^{2}+b x+c$, for which $a<b$ and $f(x) \geq 0$ for every $x \in \mathbb{R}$. Determine the smallest possible value of the expression $A=\frac{a+b+c}{b-a}$.
Solution. First method. From $f(x) \geq 0$ for every $x \in \mathbb{R}$ it follows that $a>0$ and $b^{2}-4 a c \leq 0$, i.e., $a>0$ and $\frac{b^{2}}{a} \leq c$. From the condition $a0$, therefore $a, b, c>0$. Furthermore, we have

$$
A=\frac{a+b+c}{b-a} \geq \frac{a+b+\frac{b^{2}}{4 a}}{b-a}=\frac{4 a^{2}+4 a b+b^{2}}{4 a(b-a)}
$$

If we set $b-a=y>0$, we get $b=a+y$, so

$$
A \geq \frac{4 a^{2}+4 a(a+y)+(a+y)^{2}}{4 a y}=\frac{9 a^{2}+6 a y+y^{2}}{4 a y}=\frac{3}{2}+\frac{1}{2 a y} \cdot \frac{9 a^{2}+y^{2}}{2} \geq \frac{3}{2}+\frac{1}{2 a y} \sqrt{9 a^{2} y^{2}}=3
$$

The smallest value that $A$ can achieve is $A=3$ and it is obtained for

$$
9 a^{2}=y^{2} \text { and } c=\frac{b^{2}}{4 a} \text {, i.e., } y=3 a, b=a+y=4 a \text { and } c=4 a \text {. }
$$

Thus, $A_{\min }=3$ for $b=c=4 a$.

Second method. As in the first method of solving, we conclude $a>0$ and $\frac{b^{2}}{a} \leq c$. Let $t=\frac{b}{a}$. Clearly, $t>1$. We have

$$
A=\frac{a+b+c}{b-a}=1+\frac{2 a+c}{b-a} \geq 1+\frac{2 a+\frac{b^{2}}{4 a}}{b-a}=1+\frac{8+t^{2}}{4(t-1)}=1+\frac{(t-4)^{2}+8(t-1)}{4(t-1)}=3+\frac{(t-4)^{2}}{4(t-1)} \geq 3
$$

where the smallest possible value is 3 and it is achieved for $t=4$. According to this, $A_{\min }=3$ for $b=c=4 a$.