三角函数不等式题
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三角函数不等式题
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![三角函数不等式题](/uploads/image/z/9204953-41-3.jpg?t=%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E4%B8%8D%E7%AD%89%E5%BC%8F%E9%A2%98)
由A+B+C = π,C/2 = π/2-(A/2+B/2).
tan(C/2) = 1/tan(A/2+B/2) = (1-tan(A/2)·tan(B/2))/(tan(A/2)+tan(B/2)).
乘开即得等式tan(A/2)·tan(B/2)+tan(B/2)·tan(C/2)+tan(C/2)·tan(A/2) = 1.
由不等式x²+y²+z² ≥ xy+yz+zx对任意实数x,y,z成立(可展开(x-y)²+(y-z)²+(z-x)² ≥ 0证明).
取x = tan(A/2),y = tan(B/2),z = tan(C/2)即得:
tan²(A/2)+tan²(B/2)+tan²(C/2) ≥ tan(A/2)·tan(B/2)+tan(B/2)·tan(C/2)+tan(C/2)·tan(A/2) = 1.
tan(C/2) = 1/tan(A/2+B/2) = (1-tan(A/2)·tan(B/2))/(tan(A/2)+tan(B/2)).
乘开即得等式tan(A/2)·tan(B/2)+tan(B/2)·tan(C/2)+tan(C/2)·tan(A/2) = 1.
由不等式x²+y²+z² ≥ xy+yz+zx对任意实数x,y,z成立(可展开(x-y)²+(y-z)²+(z-x)² ≥ 0证明).
取x = tan(A/2),y = tan(B/2),z = tan(C/2)即得:
tan²(A/2)+tan²(B/2)+tan²(C/2) ≥ tan(A/2)·tan(B/2)+tan(B/2)·tan(C/2)+tan(C/2)·tan(A/2) = 1.