在锐角ΔABC中，求证:
sinA+sinB+sinC+tanA+tanB+tanC>2π
证明 因为ΔABC是锐角三角形,
由万能公式得:
sinA+tanA
=2tan(A/2)/{1+[tan(A/2)]^2}+2tan(A/2)/{1-[tan(A/2)]^2}
=4tan(A/2)/{1+[tan(A/2)]^2}*{1-[tan(A/2)]^2}
运用:1/xy>=4/(x+y)^2及tanz>z，得:
sinA+tanA>4tan(A/2)>2A (1-1)
同理可得:
sinB+tanB>2B (1-2)
sinC+tanC>2C (1-3)
(1-1)+(1-2)+(1-3) 及A+B+C=π，即得:
sinA+sinB+sinC+tanA+tanB+tanC>2π.
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