命题 在任意ΔABC中,A,B,C表示其内角,求证:

命题 在任意ΔABC中,A,B,C表示其内角,求证:
(cosA)^3+(cosB)^3+(cosC)^3≥3/8.

证明(一) 由三倍角公式:
cos3A=4(cosA)^3 -3cosA (cosA)^3=(3cosA+cos3A)/4
sin3A=3sinA-4(sinA)^3=sinA*[3-4(sinA)^2]
及三角恒等式:[R,r若示外接与内切圆半径,s为半周长]
cosA+cosB+cosC=1+4sin(A/2)*sin(B/2)*sin(C/2)
cos3A+cos3B+cos3C=1-4sin(3A/2)*sin(3B/2)*sin(3C/2)
4sin(A/2)*sin(B/2)*sin(C/2)=r/R
4sin(3A/2)*sin(3B/2)*sin(3C/2)=r[3s^2-(3R+r)^2]/R^3
所以(cosA)^3+(cosB)^3+(cosC)^3
=3(cosA+cosB+cosC)/4+(cos3A+cos3B+cos3C)/4
=1+3sin(A/2)*sin(B/2)*sin(C/2)-sin(3A/2)*sin(3B/2)*sin(3C/2)
=1-5∏sin(A/2)+ [∏sin(A/2)]*{8-∏{3-4[sin(A/2)]^2}
=1-5∏sin(A/2)+ [∏sin(A/2)]*{8-∏{4[cos(A/2)]^2-1}
其中∏表示循环求积.易证
1-5∏sin(A/2)≥3/8;
8-∏{4[cos(A/2)]^2-1}≥0
所以(cosA)^3+(cosB)^3+(cosC)^3≥3/8.
证明(二) 设BC=a,CA=b,AB=c,由余弦定理得:
∑a^3*(b^2+c^2-a^2)^3≥3(abc)^3.(1)
其中∑表示循环求和.
(1)式展开整理为
-∑a^9+3∑a^7*(b^2+c^2)+∑a^6*(b^3+c^3)-3∑a^5*(b^4+c^4)-6(abc)^2*∑a^3+3(abc)^2*∑a^2*(b+c)-3(abc)^3≥0
因为a,b,c是对称的,不失一般性,设a=max(a,b,c),上式分解为
[-a^7-a^6*(b+c)+2a^5*(b^2+c^2)-a^5*bc+3a^4*(b^3+c^3)+2a^4*bc(b+c)+3a^3*bc(b^2+c^2)-a^3*b^2*c^2-3a^2*(b^5+c^5)+3a^2*b^2*c^2(b+c)-2a*(b^6+c^6)-3a*b*c*(b^4+c^4)+3a*b^2*c^2*(b^2+c^2)+4*a*b^3*c^3+2b^7+2c^7-3b*c*(b^5+c^5)-6b^2*c^2*(b^3+c^3)+7b^3*c^3*(b+c)]*(a-b)*(a-c)+(b-c)^2*{-a^2*[b^5+c^5+bc(b^3+c^3)+b^2*c^2*(b+c)]+a[2b^6+2c^6+5bc(b^4+c^4)+2b^2*c^2*(b^2+c^2)-3b^3*c^3]-[b^7+c^7+4bc(b^5+c^5)+b^2*c^2*(b^3+c^3)-9b^3*c^3*(b+c)]}≥0.
{a^2*(b+c-a)*[a^4+2a^3*(b+c)+5a^2*bc-3a(b^3+c^3)+3abc(b+c)-3bc(b^2+c^2)+b^2*c^2]+[3a^3*(b+c)^2-3a^2*(b+c)*(b^2+c^2)-a(2b^4+7b^3*c+9b^2*c^2+7b*c^3+2c^4)+2b^5+b^4*c-5b^3*c^2-5b^2*c^3+b*c^4+2c^5]*(b-c)^2+5a^2*b^2*c^2*(b+c)}*(a-b)*(a-c)
+(b-c)^2*{(b+c-a)*[a(b^5+c^5)+abc(b^3+c^3)+ab^2*c^2*(b+c)-(b^6+c^6)-bc (b^4+c^4)+b^3*c^3]+2bc(b+c)*(b-c)^2*[a(b+c)-(b^2+c^2+bc)]+7b^3*c^3*(b+c)}*(b-c)^2≥0