Question

Cho a,b,c la cac so duong sao cho a+b+c = 1 

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

Answer 1

cmtt \(\frac{b^2}{a+c}+\frac{a+c}{4}\ge b\)

\(\frac{c^2}{a+b}+\frac{a+b}{4}\ge c\)

\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}+\frac{a+b+b+c+c+a}{4}\ge a+b+c\)

\(A+\frac{1}{2}\ge1\)

Answer 2

\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2}{4}}=a\)

cmtt 

A+1/2\(\ge1\Rightarrow A\ge\frac{1}{2}\)

A là biểu thức bên trái nha