证明：
先证一个结论.
设x,y≥0且x+y≤ 2/3,则(1-x) x^2+(1-y) y^2 ≤ (1-x-y) (x+y)^2.
(1-x) x^2+(1-y) y^2-(1-x-y) (x+y)^2
=[(1-x) x^2-(1-x-y)x^2]+[(1-y) y^2-(1-x-y)y^2]-2xy(1-x-y)
=yx^2+xy^2-2xy(1-x-y)
=xy[3(x+y)-2]
≤0
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下面用数学归纳法来证明原命题.
当n=2时,容易验证结论是正确的.
设当n=m (m≥2)时,原命题是正确的.
当n=m+1时,不失一般性,设X1≥X2≥X3≥.≥Xm≥ Xm+1,则Xm+Xm+1≤2/(m+1)≤2/3.
(若Xm+ Xm+1>2/m,则 2=2(X1+X2+...+Xm+Xm+1)=(X1+X2)+(X2+X3)+...+(Xm+Xm+1)+(Xm+1+X1)≥(m+1)(Xm+Xm+1)>2,矛盾.)
用刚开始得到的结论,显然有：(1-Xm)(Xm)^2+(1-Xm+1)(Xm+1)^2≤(1-Xm-Xm+1)(Xm+Xm+1)^2.
设Y1=X1,Y2=X2,...,Ym-1=Xm-1,Ym=Xm+Xm+1,则
X1X2(X1+X2)+...+X1Xn(X1+Xn)+X2X3(X2+X3)...XmXm+1(Xm+Xm+1)
=(1-X1)(X1)^2+(1-X2)(X2)^2+.+(1-Xm)(Xm)^2+(1-Xm+1)(Xm+1)^2
≤(1-Y1)(Y1)^2+(1-Y2)(Y2)^2+.+(1-Ym)(Ym)^2
≤1/4,
证毕.