Question

Answer 1

Từ giả thiết suy ra: \(a+b+c\ge3\sqrt[3]{abc}=3\)

Đặt vế trái là P

\(P=\dfrac{a^2}{2ab+a}+\dfrac{b^2}{2bc+b}+\dfrac{c^2}{2ca+a}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)+a+b+c}\ge\dfrac{\left(a+b+c\right)^2}{\dfrac{2}{3}\left(a+b+c\right)^2+a+b+c}\)

\(\Rightarrow P\ge\dfrac{3\left(a+b+c\right)}{2\left(a+b+c\right)+3}=\dfrac{2\left(a+b+c\right)+\left(a+b+c\right)}{2\left(a+b+c\right)+3}\ge\dfrac{2\left(a+b+c\right)+3}{2\left(a+b+c\right)+3}=1\) (đpcm)

Dấu "=" xảy  ra khi \(a=b=c=1\)