论文部分内容阅读
第十三届(1953牛)普特南数学竞赛有这样一道试题: 设实数a,b,c中任意两个之和大于第三个,求证 2/3(a+b+c)(a~2+b~2+c~2) >a~3+b~3+c~3+abc. (1) 事实上,我们有命题设实数a,b,c中任意两个之和大于第二个,则 2/3(a+b+c)(a~2+b~2+c~2) ≥a~3+b~3+c~3+3abc. (2)当且仅当a=b=c时等号成立. 证明:不难验证,(2)式等价于 (b+c-a)(c+a-b)(a+b-c)
The Putnam contest of mathematics in the 13th (1953 New Bull) has such a test: Set the sum of any two of the real numbers a, b, and c larger than the third, and verify that 2/3 (a+b+c) (a~ 2+b~2+c~2) >a~3+b~3+c~3+abc. (1) In fact, we have the proposition that the sum of any two of the real numbers a, b, and c is greater than the second One, then 2/3(a+b+c)(a~2+b~2+c~2) ≥a~3+b~3+c~3+3abc. (2)If and only if a= When b=c, the equal sign is established. Proof: It is not difficult to verify that (2) is equivalent to (b+ca)(c+ab)(a+bc)