设x,y,z∈R+,求证：2x∧2/(y+z)+2y∧2/(x+z)+2z∧2/(x+y)≥x+y+z
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证：2x∧2/(y+z)-x=x(x-y)/(y+z)+x(x-z)/(y+z),2y∧2/(x+z)-y=y(y-x)/(x+z)+y(y-z)/(y+z),2z∧2/(x+y)-z=z(z-x)/(x+y)+z(z-y)/(x+y);

x(x-y)/(y+z)+y(y-x)/(x+z)=(x-y)^2*(x+y+z)/(y+z)(x+z)≥0,同理,x(x-z)/(y+z)+z(z-x)/(x+y)≥0,y(y-z)/(y+z)+z(z-y)/(x+y)≥0,于是命题得证.

注：具体的计算过程已省略,只写出了结果,请自己验证一下.

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