# 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
4 years ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $\boxed{123}$

Sort by:

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.

- 4 years 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.

- 4 years ago

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

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

- 3 years, 12 months ago

In other words, direct AM-GM works.

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

Staff - 4 years ago

Thanks for mentioning.

- 4 years ago

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

- 4 years ago