竞赛试题:设x,y,z为正数.求证
[x(2x+y+z)]^(1/2)+[y(2y+z+x)]^(1/2)+[2(z+x+y)]^(1/2)=<2(x+y+z)

竞赛试题:设x,y,z为正数.求证
[x(2x+y+z)]^(1/2)+[y(2y+z+x)]^(1/2)+[2(z+x+y)]^(1/2)=<2(x+y+z)

题是否抄错？ [2(z+x+y)]^(1/2)？？？
看前面两项，应该为
[z(2z+x+y)]^(1/2)。

[x(2x+y+z)]^(1/2)+[y(2y+z+x)]^(1/2)+[z(2z+x+y)]^(1/2)=<2(x+y+z)


证明 因为
(y^2+z^2+4yz+xy+xz)^2-4yz(2y+z+x)*(2z+x+y)
=(x+y+z)^2*(y-z)^2≥0
故得:
y^2+z^2+4yz+xy+xz≥2√[yz(2y+z+x)*(2z+x+y)],
z^2+x^2+4zx+yz+xy≥2√[zx(2z+x+y)*(2x+y+z)],
x^2+y^2+4xy+zx+yz≥2√[xy(2x+y+z)*(2y+z+x)].

上述三式相加得:
2∑x^2+6∑yz≥2∑√[yz(2y+z+x)*(2z+x+y)],

上式两边各加2∑x^2+2∑yz,整理得:
4(x+y+z)^2≥{∑√[x(2x+y+z)]}^2
开方即得所证不等式.