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.

#Algebra

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Comments

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.

Ivan Koswara - 3 years, 10 months 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.

Steven Jim - 3 years, 10 months ago

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

Ivan Koswara - 3 years, 10 months ago

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

Jon Haussmann - 3 years, 10 months ago

In other words, direct AM-GM works.

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

Calvin Lin Staff - 3 years, 10 months ago

Thanks for mentioning.

Steven Jim - 3 years, 10 months ago

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

Pi Han Goh - 3 years, 10 months ago

Here's the link.

Steven Jim - 3 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Why is that bbb so non-trivial?

Comments

Why is that $b$ so non-trivial?