设a,b,c是正实数,求证:aabbcc≥(abc)a+b+c/3.
问题描述:
设a,b,c是正实数,求证:aabbcc≥(abc)
. a+b+c 3
答
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+algcalga+blgb+clgc≥clga+algb+blgcalga+blgb+clgc=alga+blgb+clgc上述三式相加得:3(alga+blgb+clgc)≥(a+b+c)(lga+lg...