Question

Chứng minh rằng :\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\) ( a, b > 0 )

Answer 1

Giả sử \(a\ge b\Rightarrow a=b+m\left(m\ge0\right)\)

Ta có :

\(\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{b+m}{b}+\dfrac{b}{b+m}\) \(=1+\dfrac{m}{b}+\dfrac{b}{b+m}\ge1\) \(+\dfrac{m}{b+m}+\dfrac{b}{b+m}\)

\(=1+\dfrac{m+b}{b+m}=1+1=2\)

Dấu \("="\) chỉ xảy ra khi \(\left\{{}\begin{matrix}m=0\\a=b\end{matrix}\right.\)

Vậy \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\) \(\rightarrowđpcm\)

~ Chúc bn học tốt ~

Answer 2

Ta có : \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}+\dfrac{b}{a}}\) ( theo bất đẳng thức Cô-si )

\(\Rightarrow\) \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)

Answer 3

g​iả​ sử \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\left(a,b>0\right)\)

thì \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)

\(\Leftrightarrow\dfrac{a^2+b^2}{ab}\ge\dfrac{2ab}{ab}\)

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

\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{ab}\ge0\) đ​úng​ với​ mọi a,b

Bài​ này​ dạng​ lớp 8 nên chị​ giải​ thế