Question

[NTT]

a) Cho a > b ; b > 0 . CMR: \(\frac{a}{b}+\frac{b}{a}\ge2\)

Answer 1

[_Guiltykamikk_]

Ta có :  \(\frac{a}{b}+\frac{b}{a}-2\)

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

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

\(=\frac{\left(a+b\right)^2}{ab}\ge0\)( luôn đúng với a >b > 0 )

Dấu "=" xảy ra khi :  \(a+b=0\Leftrightarrow a=-b\)

Vậy ....

Answer 2

[Pain zEd kAmi]

Easy làm luôn :)

a0 Ta có: \(\frac{a}{b}+\frac{b}{a}\ge2\Rightarrow\frac{a^2+b^2}{ab}\ge2\)

vì \(a>0;b>0\left(gt\right)\Rightarrow ab>0\)nên ta có:

\(\frac{a^2+b^2}{ab}\ge2\Leftrightarrow ab.\frac{a^2+b^2}{ab}\ge2ab\)

\(\Leftrightarrow a^2+b^2\ge2ab\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)( luôn đúng )

Vậy