Why is that bb so non-trivial?

I found this interesting (to me, at least) when I was solving this problem:

"Let a1,a2,...,an,b,ka_1,a_2,...,a_n,b,k be positive reals so that a1+a2+...+annka_1+a_2+...+a_n \le nk. Prove that: (a1+b)(a2+b)...(an+b)(k+b)n(a_1+b)(a_2+b)...(a_n+b) \le (k+b)^n ."

At first, my thought was: "Oh, so it's obvious that kna1a2...ank^n \ge a_1*a_2* ... *a_n. But what about the bb? There's no way to throw it away." And right now, I'm stuck at that exact point. Reverse Rearrangement doesn't seem to work, and there is a "seemingly" trivial solution using Buffalo Way, but I want to find a "better" way than just plug-and-bash.

So, my question is: Is there a "good-looking" way to solve the problem? If there is, what is your inspiration?

Any help would be greatly appreciated. Thanks for reading!

P/S: If you can find any counterexamples, please tell me immediately and I will delete/edit this note.

Note by Steven Jim
2 years, 2 months ago

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If you've proven kna1a2ank^n \ge a_1 a_2 \ldots a_n for any a1,a2,,an,ka_1, a_2, \ldots, a_n, k, substitute kk+bk \gets k+b and aiai+ba_i \gets a_i+b and you're done.

Another method: aim to maximize (a1+b)(a2+b)(an+b)(a_1+b) (a_2+b) \ldots (a_n+b). Prove that if ai<nk\sum a_i < nk then this product is not maximized. Prove that if there is any aiaja_i \neq a_j then this product is also not maximized. So the maximum of the product is achieved when ai=ka_i = k for all ii, done.

Ivan Koswara - 2 years, 2 months ago

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That is actually the one I've been thinking about: I can't guarantee that the substitution will fit. If I could then I think this problem would've been too easy as it's so trivial.

Steven Jim - 2 years, 2 months ago

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It works. Prove the special case kna1a2ank'^n \ge a'_1 a'_2 \ldots a'_n, and use it as a lemma.

Ivan Koswara - 2 years, 2 months ago

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By AM-GM, (a1+b)(a2+b)(an+b)n(a1+b)+(a2+b)++(an+b)n=a1+a2++ann+bk+b. \begin{aligned} \sqrt[n]{(a_1 + b)(a_2 + b) \dotsm (a_n + b)} &\le \frac{(a_1 + b) + (a_2 + b) + \dots + (a_n + b)}{n} \\ &= \frac{a_1 + a_2 + \dots + a_n}{n} + b \\ &\le k + b. \end{aligned} Therefore, (a1+b)(a2+b)(an+b)(k+b)n.(a_1 + b)(a_2 + b) \dotsm (a_n + b) \le (k + b)^n.

Jon Haussmann - 2 years, 2 months ago

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In other words, direct AM-GM works.

You might be thinking of another problem, namely (ai+b)(ain+b)n \prod ( a_i + b ) \leq ( \sqrt[n]{\prod a_i} + b ) ^n .

Calvin Lin Staff - 2 years, 2 months ago

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Thanks for mentioning.

Steven Jim - 2 years, 2 months ago

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What is "Buffalo way"? Is this some new inequality trick?

Pi Han Goh - 2 years, 2 months ago

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Here's the link.

Steven Jim - 2 years, 2 months ago

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