Question

cho a,b,c >0 chứng minh rằng \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}>=\dfrac{a+b}{b+c}+\dfrac{b+c}{a+b}+1\)

Answer 1

\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+1=\dfrac{a^2}{ab}+\dfrac{b^2}{bc}+\dfrac{c^2}{ca}+\dfrac{b^2}{b^2}\)

\(\ge\dfrac{\left(a+2b+c\right)}{ab+bc+ca+b^2}=\dfrac{\left(a+b\right)^2+2\left(a+b\right)\left(b+c\right)+\left(b+c\right)^2}{\left(a+b\right)\left(b+c\right)}\)

\(=\dfrac{\left(a+b\right)}{\left(b+c\right)}+\dfrac{\left(b+c\right)}{a+b}+2\)

\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{\left(a+b\right)}{\left(b+c\right)}+\dfrac{\left(b+c\right)}{a+b}+1\)