# Teaser Inequality

Here's a "teaser" inequality for an upcoming article I'm writing on an inequality I found that has this as an application:

Given that $a_1,a_2,\ldots a_n\ge 1$ are reals then prove that $(a_1^2-a_1+a_2)(a_2^2-a_2+a_3)\cdots (a_n^2-a_n+a_1)\ge a_1^2a_2^2\cdots a_n^2$

For now, I wish to see solutions with inequalities we currently have. Good luck :3

copy-pasted from AoPS lol Note by Daniel Liu
5 years, 2 months 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:

I'm going to bed now, so maybe I'll try this later. I's just wondering if the following manipulations help!

Define $b_i=a_i-1$ for all $i$. Then we gotta prove that

$\prod_{\text{cyc}}\left(a_1^2-a_1+a_2\right)=\prod_{\text{cyc}}\left(b_1^2+b_1+1+b_2\right)\ge\prod_{\text{cyc}}\left(b_1^2+b_1+1+b_1\right).$

Can anyone finish this from here?

- 5 years, 2 months ago

Great observation. Which inequality allows you to justify that step?

Staff - 5 years, 2 months ago

nice observation. for the n=2 case we can directly use C-S to prove it. it doewnt apply to higher cases though

- 5 years, 2 months ago

A simple induction will do it, I guess

- 5 years, 2 months ago

Why don't you try the induction? You may be surprised.

- 5 years, 2 months ago

Every set of n numbers can be arranged in ascending order.Let $A_n$ be the LHS of the inequality and let $a_1\leq a_2 ... \leq a_n$.For n=1 the inequality is obvious.To make the change from n to n+1, we see that $A_(n+1)=\frac{A_n}{a_n^2-a_n+a_1}.(a_n^2-a_n+a_{n+1}).(a_{n+1}^2-a_{n+1}+a_1)$,so we want $\frac{(a_n^2-a_n+a_{n+1})}{a_n^2-a_n+a_1}.(a_{n+1}^2-a_{n+1}+a_1)\geq a_{n+1}^2$ .Let $x=a_{n+1},y=a_n and z=a_1,x\geq y\geq z$.Then we want to prove that $(1+\frac{x-z}{y^2-y+z}).(x^2-x+z) \geq x^2$. That is equivalent to $x^2-x+z+\frac{x-z}{y^2-y+z}.(x^2-x+z) \geq x^2$.Now we can cancel x^2 and factor out x-z: $(x-z)(\frac{x^2-x+z}{y^2-y+z}-1) \geq 0$, which is obvious since $x\geq y \geq z$.

Q.E.D

Is this solution correct?And how can we prove the inequality without induction?

EDIT:I've only proved the case when the reals are in ascending order.But isn't the value of the LHS minimal when this is the case?

- 5 years, 2 months ago

I would have issues with assuming that they must have a certain order. It is (as yet) not obvious that this minimizes the LHS.

Staff - 5 years, 2 months ago

Yes, that is the main problem of my solution.I don't know if we can even fill the hole

- 5 years, 2 months ago

@Calvin Lin I can't seem to solve this using Induction by the straightforward way... Can you try this out?

- 5 years, 2 months ago

Induction would not an approach that I would think of, mainly because the "cross terms" do not result in anything nice.

If we insist on trying that, the straightforward way requires showing that

$\frac{ ( a_n^2 - a_n + a_{n+1} ) (a_{n+1}^2 - a_{n+1} + a_1 ) } { (a_n^2 - a_n + a_1 ) } \geq a_{n+1} ^2.$

This is equal to

$( a_1 - a_{n+1} ) ( a_n ^2 - a_n - a_{n+1}^2 + a_{n+1} ) ( a_n^2 - a_n + a_1 ) > 0$

which is not necessarily true.

Staff - 5 years, 2 months ago

Hi, just wondering. How do you save stuff to sets? I would press create new note, but it wouldn't save to the set. What am I doing wrong?

- 5 years, 2 months ago