$\dfrac{a+1}{b+1}+\dfrac{b+1}{c+1}+\dfrac{c+1}{a+1}\leq \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}$

Bài toán:
Cho $a, b, c$ dương. Chứng minh rằng :

$\dfrac{a+1}{b+1}+\dfrac{b+1}{c+1}+\dfrac{c+1}{a+1}\leq \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}$

Lời giải:

Giả sử $c=min\{a;b;c\}$

$VP-3=\sum \frac{a}{b}-3=\left ( \dfrac{a}{b}+\dfrac{b}{a}-2 \right )+\left ( \dfrac{c}{a}+\dfrac{b}{c}-1-\dfrac{b}{a} \right )$

$=\dfrac{(a-b)^2}{ab}+\dfrac{(c-a)(c-b)}{ca}$

Tương tự:

$VT-3=\sum \dfrac{a+1}{b+1}-3=\dfrac{(a-b)^2}{(a+1)(b+1)}+\dfrac{(c-a)(c-b)}{(a+1)(c+1)}$

$\le \dfrac{(a-b)^2}{ab}+\dfrac{(a-c)(b-c)}{ac}$ 

Suy ra $VT\le VP$ (đpcm).

Dấu "=" xảy ra khi $a=b=c$.