Question

Chứng minh rằng: Với mọi x,y,z thuộc R; ta có:

\(\sqrt{x^2+xy+y^2}+\sqrt{x^2+xz+z^2}\ge\sqrt{y^2+yz+z^2}\)

Answer 1

Ta có \(VT=\sqrt{\left(y+\dfrac{x}{2}\right)^2+\left(\dfrac{-\sqrt{3}x}{2}\right)^2}+\sqrt{\left(y+\dfrac{z}{2}\right)^2+\left(\dfrac{-\sqrt{3}z}{2}\right)^2}\)
Trên mặt phẳng tọa độ \(Oxy\) lấy 2 điểm
\(A\left(y+\dfrac{x}{2};\dfrac{-\sqrt{3}}{2}x\right)\)   và    \(B\left(y+\dfrac{z}{2};\dfrac{\sqrt{3}}{2}z\right)\)
\(VT=OA+OB\ge AB=\sqrt{\left(\dfrac{z-x}{2}\right)^2+\left(\dfrac{\sqrt{3}}{2}z+\dfrac{\sqrt{3}}{2}x\right)^2}\)
\(=\sqrt{x^2+xz+z^2}=VP\) (đpcm)