Question

1. Cho x,y,z > 0. Chứng minh

\(\sqrt{x^2+xy+2y^2}+\sqrt{y^2+yz+2z^2}+\sqrt{z^2+zx+2x^2}\ge2\left(x+y+z\right)\)

Answer 1

Dạo này ko tag được đâu :(

\(VT=\sum\sqrt{\frac{1}{2}\left(x+y\right)^2+\frac{1}{2}\left(x^2+y^2\right)+y^2}\ge\sum\sqrt{\frac{1}{2}\left(x+y\right)^2+\frac{1}{4}\left(x+y\right)^2+y^2}\)

\(VT\ge\sum\sqrt{\frac{3}{4}\left(x+y\right)^2+y^2}\ge\sqrt{\frac{3}{4}\left(2x+2y+2z\right)^2+\left(x+y+z\right)^2}\)

(Mincopxki)

\(\Rightarrow VT\ge\sqrt{4\left(x+y+z\right)^2}=2\left(x+y+z\right)\)

Answer 2

@Nguyễn Việt Lâm