不妨设a≥b≥c>0,则lga≥lgb≥lgc,据排序不等式,可得三个不等式,相加,即可得出结论.
证明:不妨设a≥b≥c>0,则lga≥lgb≥lgc.
据排序不等式有:
alga+blgb+clgc≥blga+clgb+algc
alga+blgb+clgc≥clga+algb+blgc
alga+blgb+clgc=alga+blgb+clgc
上述三式相加得:
3(alga+blgb+clgc)≥(a+b+c)(lga+lgb+lgc)
即lg(aabbcc)≥lg(abc)
故aabbcc≥(abc).