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證明:∵a−ca−b+a−cb−c=a−b+b−ca−b+a−b+b−cb−c=2+b−ca−b+a−bb−c≥2+2b−ca−b×a−bb−c=4,(a>b>c)∴a−ca−b+a−cb−c≥4∴1a−b+1b−c≥4a−c
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