x²=ma＋nb=(ma＋nb)(m＋n)=m²a＋n²b＋mn(a＋b)y²=m²a＋n²b＋2mn√(ab)则：x²－y²=mn[a＋b－2√(ab)]=mn[√a－√b]²显然，有：x²≥y²，即：x≥y
x²=ma＋nb=(ma＋nb)(m＋n)=m²a＋n²b＋mn(a＋b)y²=m²a＋n²b＋2mn√(ab)x²－y²=mn[a＋b－2√(ab)]=mn[√a－√b]²x²≥y²x≥y
