证明：在三角形ABC，cosA+cosB+cosC大于等于3/2
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证明一
（逐步调整法）由和差化积公式得
cosa+cosb+cosc+cos(π/3)
=2cos[(a+b)/2]cos[(a-b)/2]+2cos[(c+π/3)/2]cos[(c-π/3)/2]
<=2{cos[(a+b)/2]+cos[(c+π/3)/2]}
=4cos[(a+b+c+π/3)/4]cos[(a+b-c-π/3)/4]
<=4cos[(a+b+c+π/3)/4]
=4cos[(π+π/3)/4]
=4cos(π/3),
所以
cosa+cosb+cosc<=3cos(π/3)=3/2.
（逐步调整法）由和差化积公式得
cosa+cosb+cosc+cos(π/3)
=2cos[(a+b)/2]cos[(a-b)/2]+2cos[(c+π/3)/2]cos[(c-π/3)/2]
<=2{cos[(a+b)/2]+cos[(c+π/3)/2]}
=4cos[(a+b+c+π/3)/4]cos[(a+b-c-π/3)/4]
<=4cos[(a+b+c+π/3)/4]
=4cos[(π+π/3)/4]
=4cos(π/3),
所以
cosa+cosb+cosc<=3cos(π/3)=3/2.
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