Question

Answer 1

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

Tương tự: \(\sqrt{x^2-xz+z^2}\ge\dfrac{1}{2}\left(x+z\right)\)

Cộng vế với vế suy ra \(VT\ge x+\dfrac{1}{2}y+\dfrac{1}{2}z=\dfrac{2x+y+z}{2}=1\) (đpcm)

Dấu bằng xảy ra khi \(x=y=z=\dfrac{1}{2}\)