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
3 years, 5 months ago

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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 - 3 years, 5 months ago

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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 - 3 years, 5 months ago

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

Calvin Lin Staff - 3 years, 5 months ago

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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 - 3 years, 4 months ago

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

Daniel Liu - 3 years, 5 months ago

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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 - 3 years, 5 months ago

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A simple induction will do it, I guess

Bogdan Simeonov - 3 years, 5 months ago

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

Daniel Liu - 3 years, 5 months ago

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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 - 3 years, 5 months ago

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@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 - 3 years, 5 months ago

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

Bogdan Simeonov - 3 years, 5 months ago

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