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證明:a2b+b2a−(a+b)=a3+b3−a2b−ab2ab=a2(a−b)−b2(a−b)ab=(a2−b2)(a−b)ab=(a+b)(a−b)2ab.∵a、b都是正數,∴a+b>0,ab>0,又由a≠b,可知(a-b)2>0,故(a+b)(a−b)2ab>0,即a2b+ b2a−(a+b)>0,∴a2b…
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