Giả sử \(a\ge b\) suy ra a = b +m (m \(\ge\) 0)
Ta có \(\frac{a}{b}+\frac{b}{a}=\frac{b+m}{b}+\frac{b}{b+m}\)
\(=\frac{b}{b}+\frac{m}{b}+\frac{b}{b+m}=1+\frac{m}{b}+\frac{b}{b+m}\)
Vì \(m,b\ge0\) nên \(\frac{m}{b}\ge\frac{m}{b+m}\)
Do đó \(1+\frac{m}{b}+\frac{b}{b+m}\ge1+\frac{m}{b+m}+\frac{b}{b+m}=1+\frac{m+b}{b+m}=1+1=2\)
Vậy \(\frac{a}{b}+\frac{b}{a}\ge2\) (dấu = xảy ra \(\Leftrightarrow\) m = 0 \(\Leftrightarrow\) a =b)
Áp dụng bđt AM - GM ta có :
\(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}=2\sqrt{1}=2\)
Dấu "=" xảy ra <=> \(a=b\)