Problem 191

(1) Since $a > 0$, $b > 0$, and $a \neq b$, we have
\begin{align*}
a + b &= (a + b)\left(\dfrac{1}{a} + \dfrac{1}{b}\right) \\
&= 1 + 1 + \dfrac{b}{a} + \dfrac{a}{b} \\
&> 2 + 2\sqrt{\dfrac{b}{a} \cdot \dfrac{a}{b}} \\
&= 2 + 2(1) \\
&= 4.
\end{align*}
Hence, $a + b > 4\,$. \boxed{a + b > 4}

(2) Since $a > b > c$ and $a + b + c = 0$, we know that $a > 0$ and $c 0 \\
(a - c)^2 &> 0.
\end{align*}
Now, we factor the left side to get $(a - c)(2a + c) > 0$, noticing that $(a - c) > 0$ because $a > c$, and $2a + c = a + c + a = a - b > 0$ since $a > b$.

So, since both $(a - c)$ and $(2a + c)$ are positive, it follows that their product is also positive. Therefore, the original inequality holds true. \boxed{\dfrac{\sqrt{b^2 - ac}}{a} < \sqrt{3}}