己知a,b,c,d是正实数，且a^2/(1+a^2)+b^2/(1+b^2)+c^2/(1+c^2)+d^2/(1+d^2)=1。
求证:abcd≤1/9.
证明 设a=tanp，b=tanq，c=tanr，d=tant，
则a^2/(1+a^2)=(sinp)^2，b^2/(1+b^2)=(sinq)^2，c^2/(1+c^2)=(sinr)^2，d^2/(1+d^2)=(sint)^2，,0<p,q,r,t<π/2。那么有
(sinp)^2+(sinq)^2+(sinr)^2+(sint)^2=1，
(sinp)^2+(sinq)^2+(sinr)^2=(cost)^2，
(sinp)^2+(sinq)^2+(sint)^2=(cosr)^2,
(sinp)^2+(sinr)^2+(sint)^2=(cosq)^2,
(sinr)^2+(sinq)^2+(sint)^2=(cosp)^2,
据均值不等式
3*[(sinp)^2*(sinq)^2*(sinr)^2]^(1/3)≤(sinp)^2+(sinq)^2+(sinr)^2=(cost)^2, (1)
3*[(sinp)^2*(sinq)^2*(sint)^2]^(1/3)≤(sinp)^2+(sinq)^2+(sint)^2=(cosr)^2, (2)
3*[(sinp)^2*(sinr)^2*(sint)^2]^(1/3)≤(sinp)^2+(sinr)^2+(sint)^2=(cosq)^2, (3)
3*[(sinr)^2*(sinq)^2*(sint)^2]^(1/3)≤(sinr)^2+(sinq)^2+(sint)^2=(cosp)^2, (4)
由(1)*(2)*(3)*(4)得:
81*(sinp)^2*(sinq)^2*(sinr)^2*(sint)^2≤(cost)^2*(cosr)^2*(cosq)^2*(cosp)^2
即 (tanp)^2*(tanq)^2*(tanr)^2*(tant)^2≤1/81，
所以abcd≤1/9. 得证。
