已知：a>1,b>1,c>1,lga+lgb=1,求log(a)(c)+log(b)(c)≥4lgc.注：log(a)(c)表示以_数学解答

已知：a>1,b>1,c>1,lga+lgb=1,求log(a)(c)+log(b)(c)≥4lgc.注：log(a)(c)表示以a为底c的对数.

人气：432 ℃　时间：2020-04-08 13:12:22

解答

lga+lgb=1ab=10logaC+logbC=lgc(1/lga +1/lgb)=lgc[(lga+lgb)/(lga*lgb)]=lgc/(lga*lgb)lga+lgb=1>=2√(lga*lgb)所以1/(lga*lgb)>=4当且仅当lga=lgb时取得等号,这时候a=b=5logaC+logbC=lgc/(lga*lgb)>=4lgc...