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已知a、b、c、d、e是实数且满足a+b+c+d+e=8,a~2+b~2+c~2+d~2+e~2=16,试确定e的最大值。(美国第七届中学数学竞赛题) 解法一:判别式法 a+b+c+d+e=8 (1) a~2+b~2+c~2+d~2+e~2=16 (2)消去a得2b~2-2(8-c-d-e)b+(8-c-d-e)~2 +c~2+d~2+e~2-16=0因为b∈R,所以 (?)_1=4(8-c-d-e)~2-8[(8-c-d-e)~2 +c~2+d~2+e~2-16]≥0即3c~2-2(8-d-e)c+[(8-d-e)~2 -2(16-d~2-e~2)]≤0由于c∈R,因而关于c的二次函数的图象与x轴相交,所以 (?)=4(8-d-e)~2-12[(8-d-e)~2 -2(16-d~2-e~2)]≥0即4d~2-2(8-e)d+(8-e)~2-3(16-e~2)≤0又因d∈R,故关于d的二次函数图象与x轴相交,所以
It is known that a, b, c, d, and e are real numbers and satisfy a+b+c+d+e=8, a~2+b~2+c~2+d~2+e~2=16. Determine the maximum value of e. (The 7th Middle School Mathematical Contest in the United States) Solution One: Discriminant Method a+b+c+d+e=8 (1) a~2+b~2+c~2+d~2+e~2= 16 (2) The elimination of a is 2b~2-2(8-cde)b+(8-cde)~2 +c~2+d~2+e~2-16=0 because b∈R, so (?) _1=4(8-cde)~2-8[(8-cde)~2 +c~2+d~2+e~2-16]≥0 is 3c~2-2(8-de)c+[ (8-de)~2 -2 (16-d~2-e~2)] ≤ 0 Since c ∈ R, the image of the quadratic function for c intersects the x-axis, so (?) = 4 ( 8-de)~2-12[(8-de)~2 -2(16-d~2-e~2)]≥0 means 4d~2-2(8-e)d+(8-e)~ 2-3 (16-e~2) ≤ 0 and because of d∈R, the quadratic function image of d intersects with the x axis, so