已知0a1,x^2+y=0.证明:loga(a^x+a^y)≤loga2+1/8.
证明: ∵00,a^y>0. 依均值不等式,得 a^x+a^y≥2根(a^x*a^y)=2a^[(x+y)/2]. 故log(a^x+a^y) ≤log[2a^((x+y)/2)] =log2+(x+y)/2 =log2+(x-x^2)/2 =log2-1/2*(x-1/2)^2+1/8 ≤log2+1/8. 证毕.
答:证:x^2+y=0--->y=-x^2 因此a^x+a^y>=2(a^x*a^y)=2a^[(x+y)/2]=2a^[(-x^2+x)/2](当仅当x=y---...详情>>
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