$Cho_{}$ $x,y,z_{}$ $là_{}$ $các_{}$ $số_{}$ $dương_{}.$ $C/m_{}$ $3(x^2+y^2+z^2)_{}$ $\geq$ $(x+y+z)^2_{}$
$Cho_{}$ $x,y,z_{}$ $là_{}$ $các_{}$ $số_{}$ $dương_{}.$ $C/m_{}$ $3(x^2+y^2+z^2)_{}$ $\geq$ $(x+y+z)^2_{}$
By aikhanh
By aikhanh
$Cho_{}$ $x,y,z_{}$ $là_{}$ $các_{}$ $số_{}$ $dương_{}.$ $C/m_{}$ $3(x^2+y^2+z^2)_{}$ $\geq$ $(x+y+z)^2_{}$
Đáp án:
Ta có :
`3(x^2 + y^2 + z^2) – (x + y + z)^2`
`= 3x^2 + 3y^2 + 3z^2 – x^2 – y^2 – z^2 – 2xy – 2yz – 2zx`
`= 2x^2 + 2y^2 + 2z^2 – 2xy – 2yz – 2zx`
`= (x^2 – 2xy + y^2) + (y^2 – 2yz + z^2) + (z^2 – 2zx + x^2)`
`= (x – y)^2 + (y – z)^2 + (z – x)^2 ≥ 0`
`=> 3(x^2 + y^2 + z^2) ≥ (x + y + z)^2` (`dpcm`)
Giải thích các bước giải: