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# Why is that $$b$$ so non-trivial?

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.

Note by Steven Jim
7 months, 3 weeks ago

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If 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.

- 7 months, 3 weeks ago

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.

- 7 months, 3 weeks ago

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

- 7 months, 3 weeks ago

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, $(a_1 + b)(a_2 + b) \dotsm (a_n + b) \le (k + b)^n.$

- 7 months, 2 weeks ago

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

- 7 months, 3 weeks ago

- 7 months, 3 weeks ago

In other words, direct AM-GM works.

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

Staff - 7 months, 3 weeks ago

Thanks for mentioning.

- 7 months, 3 weeks ago