在△ABC中，试确定t使下式成立:
(sinA)^t+(sinB)^t+(sinC)^t≤[cos(A/2)]^t+[cos(B/2)]^t+[cos(C/2)]^t

在△ABC中，试确定t使下式成立:
(sinA)^t+(sinB)^t+(sinC)^t≤[cos(A/2)]^t+[cos(B/2)]^t+[cos(C/2)]^t (1)
证明 经验证0<t≤2时(1)成立。下面证明如下
由于所证不等式关于A,B,C是对称的，不妨设A≥B≥C，则π/3≥C>0。
因为0<t/2≤1，运用幂平均不等式得
(sinA)^t+(sinB)^t=[(sinA)^2]^(t/2)+[(sinB)^2]^(t/2)
≤2^(1-t/2)*[(sinA)^2+(sinB)^2]^(t/2)=2^(1-t/2)*[1+cosC*cos(A-B)]^(t/2)
≤2^(1-t/2)*[1+cosC]^(t/2)=2[cos(C/2)]^t。
所以得:(sinA)^t+(sinB)^t-2[cos(C/2)]^t≤0 (2)
令(B-C)/2=x，则π/4>(B-C)/2≥0。那么
(sinC)^t+[cos(C/2)]^t-[cos(A/2)]^t-[cos(B/2)]^t
=(sinC)^t+[cos(C/2)]^t-[sin(x+C)]^t-[cos(x+C/2)]^t
记f(x)=(sinC)^t+[cos(C/2)]^t-[sin(x+C)]^t-[cos(x+C/2)]^t
将C视作常数，对函数f(x) 求导:
[f(x)]’=(t/2)*[(sin(π/2-x-C/2)]^(t-2)*sin(2x+C)-(t/2)*[(sin(x+C)]^(t-2)*sin(2x+2C).
因为 π/2>π/2-x-C/2=(A+C)/2>=(B+C)/2=x+C>0, t-2≤0,
所以 0<[(sin(π/2-x-C/2)]^(t-2)≤[(sin(x+C)]^(t-2).
因为 sin(2x+C)-sin(2x+2C)=-2cos(2x+3C/2)*sin(C/2),
0<2x+3C/2=π/2-(A-B)/2<π/2,0<C/2≤π/6,0<2x+C=B<π/2.
所以 0<sin(2x+C)≤sin(2x+2C).
因此 当x∈[0,π/4)时,[f(x)]’≤0.
即固定C(0<C≤π/3)时，f(x)在区间[0,π/4)上单调减少。
因而,当x∈[0,π/4)时,f(x)≤f(0)=0.
故(sinC)^t+[cos(C/2)]^t-[cos(A/2)]^t-[cos(B/2)]^t≤0 (3)
由(2),(3)式知(1)式成立.证毕.

设A=B=λ, C=π-2λ，代入(1)式得:
2^t*[sin(λ/2)]^t*{1+2^t*(cosλ)^t}≤2
当λ→0，t>0，当λ→π/2，t≤2，所以得:0<≤2。