Question

排序不等式.设a,b,c是正数,求证:a^ab^bc^c>等于（abc)^(a+b+c

Answer 1

a^(3a)*b^(3b)*c^(3c)/[(abc)^(a+b+c)]

=a^(2a-b-c)*b^(2b-c-a)*c^(2c-a-b)

=(a/b)^(a-b)*(b/c)^(b-c)*(c/a)^(c-a),

当a>=b>0时a/b>=1,a-b>=0,(a/b)^(a-b)>=1;

当0=1,(abc)^(a+b+c)>0,

∴a^(3a)*b^(3b)*c^(3c)>=(abc)^(a+b+c),

两边开立方,就得待证式.