1961年匈牙利数学奥林匹克试题第2题为: 由正数日1,61,f1构成的三个乘积(1一口)6,(1—6)c,(1一f)口不可能同时大一.1十百’ 将这道流传甚广的赛题推广,笔者新近发现了以下 定理 若口,6,c∈(O,1),“∈N。,则(1一n)护,(1—6),,(1一c)口”不可能同时大=} 垡。(”+1)“’、’证明 The 2nd question of the 1961 Hungarian Mathematical Olympiad Test Paper is: The three products (11) 6, (1–6)c consisting of 1,61,f1 on the positive day, (1−f) can not be both freshmen. One hundred and 100 ’to promote this widely popular game title, the author recently found the following theorem, 6, c ∈ (O, 1), ” ∈ N., then (1 a n) protection, (1 -6),, (1 - c) “Impossible not to be large at the same time =} 垡. (”+1)“,” Proof