Help mình chứng minh bài này nhé

[light_future96]

Cho a,b,c > 0, chứng minh rằng
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a+b}{b+c}+\frac{b+c}{a+b}+1\]

 

[bomkute1996th]

\[\]Ta có:

Sau đó áp dụng Cô-si là ra
\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a+b}{b+c}+\frac{b+c}{a+b}+1

\Leftrightarrow \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{{a+b}^{2}+{b+c}^{2}+(a+b)(b+c){(b+c)(a+b)}

\Leftrightarrow \frac{a(b+c)(a+b)}{b}+\frac{b(b+c)(a+b)}{c}+\frac{c(a+b)(b+c)}{a}\geq {a+b}^{2}+{b+c}^{2}+(a+b)(b+c)

\Leftrightarrow \frac{a(ab+ac+{b}^{2}}{b}+\frac{b(ab+{b}^{2}+ac+bc)}{c}+\frac{c(ab+{b}^{2}+ac+bc}{a}\geq {a}^{2}+3{b}^{2}+{c}^{2}+3ab+3bc+ac

\Leftrightarrow \frac{{a}^{2}c}{b}+\frac{{b}^{2}a}{c}+\frac{{b}^{3}}{c}+\frac{{b}^{2}c}{a}+\frac{{c}^{2}b}{a}\geq 2{b}^{2}+ab+2bc