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

#Algebra #Inequality #Product #Induction #Original

Easy Math Editor

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Comments

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?

Jubayer Nirjhor - 6 years, 7 months ago

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

Calvin Lin Staff - 6 years, 7 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

Daniel Liu - 6 years, 7 months ago

A simple induction will do it, I guess

Bogdan Simeonov - 6 years, 7 months ago

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

Daniel Liu - 6 years, 7 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?

Bogdan Simeonov - 6 years, 7 months ago

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

Calvin Lin Staff - 6 years, 7 months ago

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

Bogdan Simeonov - 6 years, 7 months ago

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

Daniel Liu - 6 years, 7 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.

Calvin Lin Staff - 6 years, 7 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?

Nolan H - 6 years, 6 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}`	$\boxed{123}$