Question

Cho 0<a≤b≤c .chứng minh rằng

\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\)≥\(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\)

Answer 1

Lời giải:
Xét hiệu:

\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\)

\(=\frac{a-c}{b}+\frac{b-a}{c}+\frac{c-b}{a}=-\frac{(b-a)+(c-b)}{b}+\frac{b-a}{c}+\frac{c-b}{a}\)

\(=\frac{b-a}{c}-\frac{b-a}{b}+\frac{c-b}{a}-\frac{c-b}{b}\)

\(=(b-a)(\frac{1}{c}-\frac{1}{b})+(c-b)(\frac{1}{a}-\frac{1}{b})\)

\(=\frac{(b-a)(b-c)}{bc}+\frac{(c-b)(b-a)}{ab}=(b-a)(b-c)(\frac{1}{bc}-\frac{1}{ab})\)

\(=\frac{(b-a)(b-c)(a-c)}{abc}\geq 0\) do \(0\leq a\leq b\leq c\)

Do đó:
\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{b}{a}+\frac{c}{b}+\frac{a}{c}\) (đpcm)

Dấu bằng xảy ra khi $a=b=c$