Question

Cho các số nguyên a,b và a>b>0 .Hãy chứng tỏ:

\(\frac{a}{b}+\frac{b}{a}\ge2\)

Answer 1

Xét hiệu :

\(\frac{a}{b}+\frac{b}{a}-2=\frac{a^2}{ab}+\frac{b^2}{ab}-\frac{2ab}{ab}=\frac{a^2-2ab+b^2}{ab}\)

\(=\frac{a^2-ab-ab+b^2}{ab}=\frac{a\left(a-b\right)-b\left(a-b\right)}{ab}\)\(=\frac{\left(a-b\right)\left(a-b\right)}{ab}\)\(=\frac{\left(a-b\right)^2}{ab}\)

Vì \(\left(a-b\right)^2\ge0\) và \(ab>0\)( do a, b > 0 )

\(\Rightarrow\frac{\left(a-b\right)^2}{ab}>0\)

\(\Rightarrow\frac{a}{b}+\frac{b}{a}-2\ge0\)

Hay \(\frac{a}{b}+\frac{b}{a}\ge2\)\(\left(đpcm\right)\)

Answer 2

Ta có: \(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}=2\left(đpcm\right)\)