<s>[INST] If $ab>0$, show that $\dfrac{a}{b}+\dfrac{b}{a}\ge2$. 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.
 [/INST] 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 $\blacksquare$
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