Question

[Tạ Nguyễn Thế An]

Chứng minh nếu ba số dương x,y,z thỏa mãn : 2x+y+z=2 thì

\(\sqrt{x^2-xy+y^2}+\sqrt{x^2-xz+z^2}\ge1\)

Answer 1

[Đoàn Đức Hà]

\(x^2-xy+y^2=\frac{1}{4}\left(x+y\right)^2+\frac{3}{4}\left(x-y\right)^2\)

\(\Rightarrow\sqrt{x^2-xy+y^2}=\sqrt{\frac{1}{4}\left(x+y\right)^2+\frac{3}{4}\left(x-y\right)^2}\ge\sqrt{\frac{1}{4}\left(x+y\right)^2}=\frac{1}{2}\left(x+y\right)\)

Tương tự ta cũng có: \(\sqrt{x^2-xz+z^2}=\frac{1}{2}\left(x+z\right)\)

Suy ra \(\sqrt{x^2-xy+y^2}+\sqrt{x^2-xz+z^2}\ge\frac{1}{2}\left(2x+y+z\right)=1\)

Dấu \(=\)xảy ra khi \(x=y=z=\frac{1}{2}\).