【急】设a,b,c均为大于1的正数,且ab=10,求证:loga(c)+logb(c)>=4lgc如题1ga*lgb

【急】设a,b,c均为大于1的正数,且ab=10,求证:loga(c)+logb(c)>=4lgc如题1ga*lgb
【急】设a,b,c均为大于1的正数,且ab=10,求证:loga(c)+logb(c)>=4lgc
如题
1ga*lgb

【急】设a,b,c均为大于1的正数,且ab=10,求证:loga(c)+logb(c)>=4lgc如题1ga*lgb
loga(c)+logb(c)=lgc/lga+lgc/lgb=lgc
=lgc =lgc(1/(lga*lgb) 1ga*lgb =4lgc
得证

loga(c)+logb(c)=lgc/lga+lgc/lgb
=lgc[(lgb+lga)/(lga*lgb)]
=lgc[(lgab/(lga*lgb)] =lgc(1/(lga*lgb)
(lga-lgb)^2 >=0
[(lga)^2+(lgb)^2]/2 >= lga...

全部展开

loga(c)+logb(c)=lgc/lga+lgc/lgb
=lgc[(lgb+lga)/(lga*lgb)]
=lgc[(lgab/(lga*lgb)] =lgc(1/(lga*lgb)
(lga-lgb)^2 >=0
[(lga)^2+(lgb)^2]/2 >= lga*lgb
1/2=(lga+lgb)^2 /2 >=[(lga)^2+(lgb)^2]/2 >= lga*lgb
1/(lga*lgb)>=1/(1/2)=2 (因 lga>=0,lgb>=0)
loga(c)+logb(c) =lgc(1/(lga*lgb) >= 2lgc

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