Question

Cho \(0< a\le b\le c.CMR:\frac{b}{c}+\frac{c}{a}\ge\frac{b}{a}+\frac{a}{b}\)

Answer 1

Lời giải:

Xét hiệu:

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

\(=\frac{ab^2+bc^2-b^2c-a^2c}{abc}\geq \frac{a^2b+bc^2-b^2c-a^2c}{abc}=\frac{a^2(b-c)-bc(b-c)}{abc}=\frac{(a^2-bc)(b-c)}{abc}\)

Vì $0< a\leq b\leq c\Rightarrow a^2-bc\leq 0; b-c\leq 0$

$\Rightarrow \frac{b}{c}+\frac{c}{a}-\left(\frac{b}{a}+\frac{a}{b}\right)\geq 0$

$\Rightarrow \frac{b}{c}+\frac{c}{a}\geq \frac{b}{a}+\frac{a}{b}$ (đpcm)