※ 引述《adow (天下人間)》之銘言:
: 已知a,b,c>0
: 證明:√ab(a+b) + √bc(b+c) + √ca(c+a) > √(a+b)(b+c)(c+a)
: 謝謝
without loss of generality,
we may assume a >= b >= c = 1,
then we will prove
√[ab(a+b)] + √[b(1+b)] + √[a(1+a)]
> √[(a+b)(1+b)(1+a)]
proof. first, we see that
√[b(1+b)] > b, √[a(1+a)] > a.
hence,
√[ab(a+b)] + √[b(1+b)] + √[a(1+a)]
> √[ab(a+b)] + b + a = √(a+b) [√(a+b) + √(ab)]
next, consider
[√(a+b) + √(ab)]^2 = a+b+ab + 2√[(a+b)ab]
> a+b+ab+1 = {√[(1+b)(1+a)] }^2
done.
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新梗題 good question
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