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91年高中数学联赛第一试第五大题是:已知00,a~y>0,有: a~x+a~y≥(a~x·a~y)~(1/2)=2a((x+y)/2),从而 log_a(a~x+a~y)≤log_a~(2a((x+y)/2)) (1) 现有log_a(2a((x+y)/2)) =log_a2+(x+y)/2=log_a2+(1/2)x(1-x) ≤log_a2+(1/2)·(1/2)~2+log_a~2+1/8 (2) 命题得证。下面我们来证明上述(1)式与(2)式中的等号不可能同时成立,亦即题目中的结论不能取等号: (1)式中等号成立等价于χ=y,而(2)式中等号成立等价于χ=(1/2)。故(1)、(2)两式等号同时成立时必有
The fifth topic of the 91st high school mathematics league match is: Know that 00,a~y>0, there are: a~x+a~y≥(a~x•a~y)~ (1/2)=2a((x+y)/2), so log_a(a~x+a~y)≤log_a~(2a((x+y)/2)) (1) Existing log_a ( 2a((x+y)/2)) =log_a2+(x+y)/2=log_a2+(1/2)x(1-x) ≤log_a2+(1/2)·(1/2)~2+log_a ~2+1/8 (2) The proof of the proposition. Let us prove that the equal signs in (1) and (2) above cannot be established at the same time. That is, the conclusion in the title cannot be equal to: (1) The equal sign is equivalent to χ=y, and ( 2) The equal sign is equivalent to χ=(1/2). Therefore, the two (1) and (2) equal signs must be established at the same time.