logac=lgc/lga,logbc=lgc/lgb
logac+logbc
=lgc/lga+lgc/lgb
=(lgclgb+lgclga)/(lgalgb)
=lgc(lgb+lga)/(lgalgb)因为ab=10,所以lga+lgb=1
=lgc/lgalgb
因为ab=10,两边同时取对数,得
lgab=lg10
lga+lgb=1两边平方,得
lg^2a+2lgalgb+lg^2b=1因为lg^2a+lg^2b≥2lgalgb
所以1≥4lgalgb,即1/lgalgb≥4,两边同时乘以lgc,
lgc/lgalgb≥4lgc
结论得证.