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2008年德国数学奥林匹克有一题如下:题目求最小的常数c,使得对所有的实数x、y,有1+(x+y)~2≤c(1+x~2)(1+y~2).解法1取x=y=(?)/2,代入原不等式得3≤9/4c(?)c≥4/3.下面证明:c=4/3满足条件.只需证:1+(x+y)~2≤4/3(1+x~2)(1+y~2).化简整理得(2xy-1)~2+(x-y)~2≥0.这是显然的.故c_(min)=4/3.
In 2008, the German Mathematical Olympiad has one question as follows: The problem of seeking the smallest constant c such that 1 + (x + y) ~ 2≤c (1 + x ~ 2) (1 + y ~ 2) for all real numbers x, y The solution 1 takes x = y = (?) / 2 and substitutes the original inequality 3≤9 / 4c (?) C≥4 / 3. It is proved that c = 4/3 satisfies the condition. (x + y) ~ 2≤4 / 3 (1 + x ~ 2) (1 + y ~ 2) .This simplifies and concludes (2xy-1) ~ 2 + (xy) ~ 2≥0. So c_ (min) = 4/3.