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*

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestI'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?

Log in to reply

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

Log in to reply

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

Log in to reply

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?

Log in to reply

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

Log in to reply

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.

Log in to reply

A simple induction will do it, I guess

Log in to reply

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

Log in to reply

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?

Log in to reply

Log in to reply

Log in to reply