分类：教育/科学-理工学科-数学
问题：高中数学已知p、q、r为锐角,且(sinp)^3+(sinq)^ 
已知p、q、r为锐角,且(sinp)^3+(sinq)^3+(sinr)^3=1.
求证:(tanp)^2+(tanq)^2+(tanr)^2>=(3根3)/2.
回答：证明:
设a=sinp,b=sinq,c=sinr,
则a^3+b^3+c^3=1,
(a-a^3)^2
=1/2*2a^2*(1-a^2)*(1-a^2)
=<1/2*[(2a^2+1-a^2+1-a^2)/3]^3
故a-a^3=<2/(3根3) ......(1)
同理,
b-b^3=<2/(3根3) ......(2)
c-c^3=<2/(3根3) ......(3)
故a^2/(1-a^2)+b^2/(1-b^2)+c^2/(1-c^2)
=a^3/[a(1-a^2)]+b^3/[b(1-b^2)]+c^3/[l(1-c^2)]
>=[2/(3根3)]*(a^3+b^3+c^3)
=(3根3)/2.
注意到(tanp)^2=(sinp)^2/[1-(sinp)^2]=a^2/(1-a^2)
(tanq)^2=(sinq)^2/[1-(sinq)^2]=b^2/(1-b^2)
(tanr)^2=(sinr)^2/[1-(sinr)^2]=c^2/(1-c^2)
所以,
(tanp)^2+(tanq)^2+(tanr)^2>=(3根3)/2.
易知,上述不等式等号不能成立.