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

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

At first, my thought was: "Oh, so it's obvious that \(k^n \ge a_1*a_2* ... *a_n\). But what about the \(b\)? 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.

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## Comments

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TopNewestIf you've proven \(k^n \ge a_1 a_2 \ldots a_n\) for any \(a_1, a_2, \ldots, a_n, k\), substitute \(k \gets k+b\) and \(a_i \gets a_i+b\) and you're done.

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

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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.

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

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By AM-GM, \[ \begin{align*} \sqrt[n]{(a_1 + b)(a_2 + b) \dotsm (a_n + b)} &\le \frac{(a_1 + b) + (a_2 + b) + \dots + (a_n + b)}{n} \\ &= \frac{a_1 + a_2 + \dots + a_n}{n} + b \\ &\le k + b. \end{align*} \] Therefore, \[(a_1 + b)(a_2 + b) \dotsm (a_n + b) \le (k + b)^n.\]

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

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Here's the link.

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In other words, direct AM-GM works.

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

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Thanks for mentioning.

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