Question

已知a,b∈R⁺,求证:a^ab^b≥a^bb^a.

Answer 1

证明:=a^a-b·b^b-a=()^a-b.

①当a＞b＞0时,＞1,a-b＞0,

∴()^a-b＞1;

②当0＜a＜b时,0＜＜1,a-b＜0,

∴()^a-b＞1;

③当a=b时,=1,a-b=0,()^a-b=1.

总之,a^ab^b≥a^bb^a.