数学lg怎么算,高2了~TOY数学专问3
摘要因为lg[(a+b)/2]≥lg√ab=(1/2)lgab=(1/2)(lga+lgb)lg[(b+c)/2]≥lg√bc=(1/2)lgbc=(1/2)(lgb+lgc),lg[(c+a)/2]≥l
因为lg[(a+b)/2]≥lg√ab=(1/2)lgab=(1/2)(lga+lgb)
lg[(b+c)/2]≥lg√bc=(1/2)lgbc=(1/2)(lgb+lgc),
lg[(c+a)/2]≥lg√ac=(1/2)lgac=(1/2)(lga+lgc),
所以lg[(a+b)/2]+lg[(b+c)/2]+lg[(a+c)/2]≥(1/2)(lga+lgb+lgb+lgc+lga+lgc)=
lfa+lgb+lgc,得证
