chứng minh rằng: x^2+ y^2+ z^2 >hoặc= 2xy – 2xz + 2xyz 04/09/2021 Bởi Lyla chứng minh rằng: x^2+ y^2+ z^2 >hoặc= 2xy – 2xz + 2xyz
Ta có ` x^2 + y^2 + z^2 \geq 2xy – 2xz + 2yz` ` => x^2 + y^2 + z^2 – 2xy + 2xz – 2yz \geq 0` ` => x(x-y-z) – y(x-y-z) + z(x-y+z) \geq 0` ` => (x-y+z)(x-y+z)^2 \geq 0` ` => ( x – y + z)^2 \geq 0` ( đpcm ) Bình luận
`x^2+y^2+z^2≥2xy-2xz+2yz`
`⇔x^2+y^2+z^2-2xy+2xz-2yz≥0`
`⇔(x-y+z)^2≥0` (luôn đúng)
`⇒đpcm`
Ta có
` x^2 + y^2 + z^2 \geq 2xy – 2xz + 2yz`
` => x^2 + y^2 + z^2 – 2xy + 2xz – 2yz \geq 0`
` => x(x-y-z) – y(x-y-z) + z(x-y+z) \geq 0`
` => (x-y+z)(x-y+z)^2 \geq 0`
` => ( x – y + z)^2 \geq 0`
( đpcm )