Question

设a,b,c是正实数，求证：a^ab^bc^c≥(abc).

Answer 1

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.

且alga+blgb+clgc=alga+blgb+clgc,

以上三式相加整理,得3(alga+blgb+clgc)≥(a+b+c)(lga+lgb+lgc),

即lg(a^ab^bc^c)≥·lg(abc).

故a^ab^bc^c≥(abc).