Question

Cho a+b+c =1. cmr \(\sqrt{2a^2+ab+2b^2}+\sqrt{2b^2+2bc+2c^2}+\sqrt{2c^2+ac+2a^2}>=\sqrt{5}\)

Answer 1

\(\sqrt{2a^2+ab+2b^2}=\sqrt{\frac{5}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2}\ge\sqrt{\frac{5}{4}\left(a+b\right)^2}=\frac{\sqrt{5}\left(a+b\right)}{2}\)

Tương tự:\(\sqrt{2b^2+bc+2c^2}\ge\frac{\sqrt{5}\left(b+c\right)}{2}\);\(\sqrt{2c^2+ca+2a^2}\ge\frac{\sqrt{5}\left(c+a\right)}{2}\)

Cộng theo vế 3 BĐT trên ta có:\(VT\ge\frac{\sqrt{5}\left(2a+2b+2c\right)}{2}=\sqrt{5}\left(a+b+c\right)=\sqrt{5}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)