HSC 2013 MX2 Marathon (archive) (8 Viewers)

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RealiseNothing

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Re: HSC 2014 4U Marathon



AM-GM will lead to an inequality that is not tight enough to cover the bound in question
AM-GM actually works, try it again (you need more than one application though).
 

RealiseNothing

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Re: HSC 2014 4U Marathon

Demonstration?
AM-GM numerator, AM-GM denominator, combine the results.

Wait I think the inequality sign is the wrong way around, will check after this LoL match (posting during deaths lel)
 

Sy123

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Re: HSC 2014 4U Marathon

AM-GM numerator, AM-GM denominator, combine the results.

Wait I think the inequality sign is the wrong way around, will check after this LoL match (posting during deaths lel)
Yea you come across the same problem that I had 30 or so pages before, where you cannot divide inequalities side by side, each AM-GM inequality you are talking about are on the opposite sides
 

RealiseNothing

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Re: HSC 2014 4U Marathon

Yea you come across the same problem that I had 30 or so pages before, where you cannot divide inequalities side by side, each AM-GM inequality you are talking about are on the opposite sides
Yep when I quickly did it I assumed it would work without checking the direction of the inequality :/
 

seanieg89

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Re: HSC 2014 4U Marathon

The question smells like some clever application of AM-GM could do it, will have a look for such a solution tomorrow when feeling less shit.
 

seanieg89

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Re: HSC 2014 4U Marathon

Yeah, had a good look for about 30mins just now. Couldn't find any solution just using AM-GM (although I still think there must be one), so my preferred solution remains Jensen's.

Otherwise things like Lagrange multipliers or an argument based on peturbation would come out pretty quickly.
 

seanieg89

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Re: HSC 2014 4U Marathon



Hint: Try to exploit symmetry.
 

seanieg89

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Re: HSC 2014 4U Marathon

Show that xyz/(x^3 + y^3 + xyz) + xyz/(y^3 + z^3 + xyz) + xyz/(z^3 + x^3 + xyz) ≤ 1 for all positive real x, y, z.
 

RealiseNothing

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Re: HSC 2014 4U Marathon

Show that xyz/(x^3 + y^3 + xyz) + xyz/(y^3 + z^3 + xyz) + xyz/(z^3 + x^3 + xyz) ≤ 1 for all positive real x, y, z.
First I will prove a simple statement. If:





So applying this to the first inequality gives:





Using the same idea as before:



So for your inequality, I will instead prove that:





Which is true as

Just going to check everything, make sure I haven't stuffed up any steps or any of the logic here.

Eh, something is wrong I'm pretty sure.
 
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Sy123

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Re: HSC 2014 4U Marathon

Show that xyz/(x^3 + y^3 + xyz) + xyz/(y^3 + z^3 + xyz) + xyz/(z^3 + x^3 + xyz) ≤ 1 for all positive real x, y, z.
Great question! Took me a while to get it, but I like my solution:



 

AAEldar

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Re: HSC 2014 4U Marathon

Can you see what is wrong with mine? I'm fairly sure there is a problem somewhere but I can't find it.
You flipped the whole left hand side without getting a common denominator first?
 

RealiseNothing

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Re: HSC 2014 4U Marathon

You flipped the whole left hand side without getting a common denominator first?
But for all 3 terms, the denominator > numerator. So when I flip it, numerator > denominator, and so it is >1?
 

RealiseNothing

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Re: HSC 2014 4U Marathon

AHHH I GOT IT.

My proof was not IFF, so I can't reverse it.
 
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