1. vì a.b>0 => [tex]\frac{a}{b}>0;\frac{b}{a}>0=>\frac{a}{b}+\frac{b}{a}\geq 2\sqrt{\frac{a}{b}.\frac{b}{a}}=2[/tex]
2. vì a.b<0 => [tex]-\frac{a}{b}>0;-\frac{b}{a}>0=>-\frac{a}{b}+(-\frac{b}{a})\geq 2\sqrt{-\frac{a}{b}.(-\frac{b}{a})}=2=>-\frac{a}{b}+(-\frac{b}{a})\geq 2<=>\frac{a}{b}+\frac{b}{a}\leq -2[/tex]
3. [tex]a+b\geq 2\sqrt{ab};\frac{1}{a}+\frac{1}{b}\geq \frac{2}{\sqrt{ab}}=>(a+b)(\frac{1}{a}+\frac{1}{b})\geq 4[/tex]
nên cả 3 đều đúng