Question

\(\dfrac{X2+Y2+Z2}{3}>=\left(\dfrac{X+Y+Z}{3}\right)^2\)

Answer 1

xét hiệu

\(\dfrac{x^2+y^2+z^2}{3}-\dfrac{\left(x+y+z\right)^2}{9}\ge0\)

<=> \(\dfrac{3\left(x^2+y^2+z^2\right)}{9}-\dfrac{x^2+y^2+z^2+2xy+2yz+2zx}{9}\ge0\)

=> \(3x^2+3y^2+3z^2-x^2-y^2-z^2-2yx-2yz-2xz\ge0\)

<=> \(2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\)

<=>\(\left(x^2-2yx+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)\ge0\)

<=> (x-y)² +(y-z)² +(x-z)² ≥ 0 (luôn đúng )

=> đpcm