在△ABC中，n≥1，求证
cos(A/n)+cos(B/n)+cos(C/n)>2+cos(π/n)

在△ABC中，n≥1，求证
cos(A/n)+cos(B/n)+cos(C/n)>2+cos(π/n).

证明 不妨设C=max(A,B,C) ，则有
cos(A/n)+cos(B/n)+cos(C/n)-2-cos(π/n)
=cos(A/n)-1+cos(B/n)-1+cos(C/n)-cos(π/n)
=-2[sin(A/2n)]^2-2[sin(B/2n)]^2+2sin[(π+C)/2n]*sin[(π-C)/2n]
故只需证
sin[(π+C)/2n]*sin[(π-C)/2n]>sin(A/2n)]^2+[sin(B/2n)]^2 (1)
若(π+C)/2n<π/2，则sin[(π+C)/2n]>sin[(π-C)/2n] (2)
若(π+C)/2n≥π/2, 则π-(π+C)/2n=[(2n-1)π-C]/2n≥(π-C)/2n>0，知(2)成立。
因此我们只需证
{sin[(π-C)/2n]}^2>sin(A/2n)]^2+[sin(B/2n)]^2 (3)
<==> [sin(A+B)/2n]^2>sin(A/2n)]^2+[sin(B/2n)]^2
而[sin(A+B)/2n]^2-sin(A/2n)]^2-[sin(B/2n)]^2
=sin(A/2n)]^2*{cos(B/2n)]^2-1}+[sin(B/2n)]^2*{[cos(A/2n)]^2-1}+2sin(A/2n)*sin(B/2n)*cos(A/2n)*cos(B/2n)
=2sin(A/2n)*sin(B/2n)*[cos(A/2n)*cos(B/2n)-sin(A/2n)*sin(B/2n)]
=2sin(A/2n)*sin(B/2n)*cos[(A+B)/2n]>0，
所以(3)式也即(1)式成立，命题得证。