己知a,b,c都是正数，abc=1。证明 1/(1+b+c)+1/(1+c+a)+1/(1+a+b)≤1/(2+a)+1/(2+b)+1/(2+c) 证明 由abc=1条件,设a=x^2/yz, b=y^2/zx, c=z^2/xy，x,y,z∈R+. 作代换,等价于 ∑xyz/(xyz+y^3+z^3)≤∑yz/(2yz+x^2) (1) 设M=1-∑xyz/(xyz+y^3+z^3),N=1-∑yz/(2yz+x^2). 则化简得: M=∑{x(y+z)*(y-z)^2/[(xyz+y^3+z^3)∑x]} N=∑{x(2xy+2xz-yz)*(y-z)^2/[2(2xy+z^2)(2xz+y^2)*∑x] 欲证(1),只需证 M≥N,即 ∑{x(y+z)*(y-z)^2/[(xyz+y^3+z^3)∑x]}≥ ∑{x(2xy+2xz-yz)*(y-z)^2/[2(2xy+z^2)(2xz+y^2)*∑x] <==> ∑{x(y+z)[6x^2*yz+2x(y^3+z^3)+yz(y^2+z^2)]+x(yz)^2∑x}(y-z)^2/[(2xy+z^2)(2xz+y^2)(xyz+y^3+z^3)]≥0. 显然成立.