Question

CMR:\(\frac{a}{b}+\frac{b}{a}\ge2\forall\frac{a}{b}\in N\)

Answer 1

\(+\)\(a=b\Rightarrow\frac{a}{b}=\frac{b}{a}=1\Rightarrow\frac{a}{b}+\frac{b}{a}=2\)

\(a< b\Rightarrow b=a+m\Rightarrow\frac{a}{b}+\frac{b}{a}=\frac{a}{a+m}+\frac{a+m}{a}\)

\(=\frac{a}{a+m}+1+\frac{m}{a}>1+\left(\frac{a}{a+m}+\frac{a}{a+m}\right)=1+1=2\)

\(\Rightarrow\frac{a}{b}+\frac{b}{a}>2\)

\(a>b\)chứng minh tương tự như với\(a< b\)

Answer 2

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

Mà \(\left(a-b\right)^2\ge0\Leftrightarrow a^2+b^2\ge2ab\)

=>\(\frac{\left(a^2+b^2\right)}{ab}\ge2\)(đpcm)