If $ab>0$, show that $\dfrac{a}{b}+\dfrac{b}{a}\ge2$.
I am really sorry for the basic question, but I want to make things clear for me. By the AM-GM inequality we have $$\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}=2$$ Is this enough? I mean we know that the inequality holds only for positive numbers $a$ and $b$, but here we can have $a<0$ and $b<0$ and then $ab>0$ will also hold.
On
I have followed another reasoning to prove the statement.
Case 1: $a=b$ $$\frac{a}{a}+\frac{a}{a}=2\quad \blacksquare$$
Case 2: $a\neq b$
Notice that $ab>0$ is the same as saying $a,b\neq 0$, and both have the same sign (which later becomes irrelevant).
$$\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}=\frac{a^2+2ab+b^2-2ab}{ab}=\frac{(a+b)^2}{ab}-2\geq2$$ $$\frac{(a+b)^2}{ab}\geq4;\quad (a+b)^2\geq 4ab;\quad a^2+2ab+b^2\geq4ab;\quad a^2-2ab+b^2=(a-b)^2\geq0$$
which we know it is true regardless of the signs of $a$, $b$ since a squared number is always positive $\blacksquare$
On
There is no need to make any cases at all. We have $ab > 0$ (this is given), $a^2 > 0$ and $b^2 > 0$. Just apply the AM-GM inequality to $\frac{a^2}{ab}$ and $\frac{b^2}{ab}$, which we know are positive numbers.
$$\frac a b + \frac b a = \frac{a^2}{ab} + \frac{b^2}{ab} \ge 2 \sqrt{\frac{a^2b^2}{a^2b^2}} = 2$$
On
The proof below works pretty well. \begin{align} &(b-a)^2\geq 0\\ \iff &a^2+b^2-2ab\geq 0\\ \iff &a^2+b^2\geq 2ab\\ \implies&\frac{a}{b}+\frac{b}{a}\geq 2. \end{align} In the last step I divided by $ab$. It works because $ab$ is positive.
You've almost had a perfect solution, but just by adding few cases, you can make it perfect.
\begin{align} &\text{Case 1. } a>0, b>0. \\ &\Rightarrow \frac a b + \frac b a \geq 2 \sqrt{\frac a b \cdot \frac b a} = 2\text{ (AM-GM).} \\ \ \\ &\text{Case 2. } a<0, b<0. \\ &\text{let } a^*=-a, b^*=-b. \Rightarrow a^*>0, b^*>0.\\ &\Rightarrow \frac a b + \frac b a=\frac {a^*} {b^*} + \frac {b^*} {a^*} \geq 2 \sqrt{\frac {a^*}{b^*} \cdot \dfrac {b^*}{a^*}}=2 \text{ (AM-GM)}. \end{align}