Question

cho các số dương a,b,c thỏa mãn a+b+c=1. chứng minh rằng \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}>=\frac{1}{2}\)

Answer 1

Cách 2 

Vì a,b,c dương nên áp dụng BĐT Cô-si ta có

\(\frac{a^2}{a+b}+\frac{a+b}{4}>=2\sqrt{\frac{a^2}{a+b}.\frac{a+b}{4}=a}\)

\(\frac{b^2}{b+c}+\frac{b+c}{4}>=2\sqrt{\frac{b^2}{b+c}.\frac{b+c}{4}=b}\)

\(\frac{c^2}{c+a}+\frac{c+a}{4}>=2\sqrt{\frac{c^2}{c+a}.\frac{c+a}{4}=c}\)

=>  \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}+\frac{2\left(a+b+c\right)}{4}>=a+b+c\)

<=> \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}>=\frac{a+b+c}{2}=\frac{1}{2}\)