Problem 222

(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.
$\text{}$
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. $\text{b 2}$ - ac}}{a} < ${\sqrt{3}}$