lg[(a+b)/2]
≥lg[2(根号ab)/2]
=lg根号ab
=lga^(1/2)+lgb^(1/2)
=1/2(lga+lgb)
=(lga+lgb)/2
命题得证看不明白lg[(a+b)/2]≥lg[2(根号ab)/2]，a+b≥2根号(ab)=lg根号ab，分子分母同时除以2=lga^(1/2)+lgb^(1/2) ，根号(ab)=(ab)^(1/2)=a^(1/2)*b^(1/2)=1/2(lga+lgb) ，lga^(1/2)=1/2lga，lgb^(1/2)=1/2lgb=(lga+lgb)/2命题得证还有哪里不懂