Question

求解一道高一数学不等式证明题

已知a,b为正数,且a>b,求证:a^a乘b^b大于(ab)^[(a+b)/2]

Answer 1

证：
(a^a×b^b)/(ab)^[(a+b)/2]
=a^[a- (a+b)/2] × b^[b-(a+b)/2]
=a^[(a-b)/2]×b^[/(b-a)/2]
=(a/b)^[(a-b)/2]
a>b>0，a/b>1 a-b>0 (a-b)/2>0
(a/b)^[^(a-b)/2]>1^0
(a/b)^[^(a-b)/2]>1
(a^a×b^b)/(ab)^[(a+b)/2]>1
(a^a×b^b)>(ab)^[(a+b)/2]

Answer 2

本题目这样思考
先用左边的除以右边的得到
a^a*b^b/(ab)^[(a+b)/2]只要它>1就行了。
下面再化简得
a^[(a-b)/2]*b^[(b-a)/2]>1即可
a^[(a-b)/2]/b^[(a-b)/2]>1
(a/b)^[(a-b)/2]>1①
a/b>1 a-b>0
得上面的①是>1的得证了