HSC 2013 MX2 Marathon (archive) (1 Viewer)

RealiseNothing · Jan 1, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Help needed$ \\ \\ $Prove$ \\ \\ \left(\frac{1-a_1}{a_1} \right) \left(\frac{1-a_2}{a_2} \right) \dots \left(\frac{1-a_n}{a_n} \right) \geq (n-1)^n \\ \\ $for positive real$ \ a_k \ $and$ \ \ a_1 + a_2 + \dots +a_n = 1$

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).

Sy123 · Jan 1, 2014

Re: HSC 2014 4U Marathon

seanieg89 said:
If you take the log of both sides, it is a straightforward application of Jensen's inequality.

Aha ok then

RealiseNothing said:
AM-GM actually works, try it again (you need more than one application though).

Demonstration?

RealiseNothing · Jan 1, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
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 · Jan 1, 2014

Re: HSC 2014 4U Marathon

RealiseNothing said:
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 · Jan 1, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
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 · Jan 1, 2014

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 · Jan 1, 2014

Re: HSC 2014 4U Marathon

asianese said:
Realise that all of $(1-a_k)$ are negative and so you can't do AM-GM to that. Pretty sure that's the issue.

Uh, no they aren't.

asianese · Jan 1, 2014

Re: HSC 2014 4U Marathon

Oh god what am I doing

seanieg89 · Jan 2, 2014

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 · Jan 5, 2014

Re: HSC 2014 4U Marathon

$Let $x,y,z\geq 0$ and let $n$ be a non-negative integer.\\ \\ Prove that:\\ \\ $x^n(x-y)(x-z)+y^n(y-x)(y-z)+z^n(z-x)(z-y)\geq 0.$$

Hint: Try to exploit symmetry.

RealiseNothing · Jan 5, 2014

Re: HSC 2014 4U Marathon

seanieg89 said:
$Let $x,y,z\geq 0$ and let $n$ be a non-negative integer.\\ \\ Prove that:\\ \\ $x^n(x-y)(x-z)+y^n(y-x)(y-z)+z^n(z-x)(z-y)\geq 0.$$

Hint: Try to exploit symmetry.

Let $x \geq y \geq z$

The last term is positive, so let's consider the first two terms:

$x^n(x-y)(x-z) + y^n(y-x)(y-z) \geq x^n(x-y)(x-z) + y^n(y-x)(x-z)$

$= (x-z)[(x^n(x-y) + y^n(y-x)]$

$= (x-z)[(x^n(x-y)-y^n(x-y)$

$= (x-z)(x-y)(x^n-y^n) \geq 0$

seanieg89 · Jan 5, 2014

Re: HSC 2014 4U Marathon

RealiseNothing said:
Let $x \geq y \geq z$

The last term is positive, so let's consider the first two terms:

$x^n(x-y)(x-z) + y^n(y-x)(y-z) \geq x^n(x-y)(x-z) + y^n(y-x)(x-z)$

$= (x-z)[(x^n(x-y) + y^n(y-x)]$

$= (x-z)[(x^n(x-y)-y^n(x-y)$

$= (x-z)(x-y)(x^n-y^n) \geq 0$

good

.

seanieg89 · Jan 5, 2014

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 · Jan 6, 2014

Re: HSC 2014 4U Marathon

seanieg89 said:
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:

$\frac{a}{d} + \frac{b}{e} + \frac{c}{f} < 1$

$\frac{a+b+c}{d+e+f} = \frac{a}{d+e+f} + \frac{b}{d+e+f} + \frac{c}{d+e+f} < \frac{a}{d} + \frac{b}{e} + \frac{c}{f}$

So applying this to the first inequality gives:

$\frac{a+b+c}{d+e+f} < 1$

$\frac{d+e+f}{a+b+c} > 1$

Using the same idea as before:

$\frac{d}{a} + \frac{e}{b} + \frac{f}{c} > 1$

So for your inequality, I will instead prove that:

$\frac{x^3 + y^3 + xyz}{xyz} + \frac{y^3 + z^3 + xyz}{xyz} + \frac{z^3 + x^3 + xyz}{xyz} \geq 1$

$\frac{2(x^3+y^3+z^3)}{xyz} + 3 \geq 1$

Which is true as $x, y, z > 0$

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.

Sy123 · Jan 6, 2014

Re: HSC 2014 4U Marathon

seanieg89 said:
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:

$\\ $Substitution$ \\ \\ a = \frac{x}{y} , b = \frac{y}{z} , c = \frac{z}{x} \\ \\ \therefore \ abc = 1 \\ \\ \sum_{cyc} \frac{xyz}{x^3 + y^3 + xyz} = \sum_{cyc} \frac{1}{1 + \frac{(x/y)^3 +1}{\frac{xz}{y^2}}} = \sum_{cyc} \frac{a}{a + b(a^3 + 1)} \\ \\ $Now consider$ \\ \\ a^2 + 1 \geq 2a \ \Rightarrow \ \ a^3 + 1 \geq a(a+1) \\ \\ \therefore \ \ \frac{a}{a + b(a^3 +1)} \leq \frac{1}{1 + b(a+1)} \\ \\ $Hence we must prove$ \ \ \ \ \sum_{cyc} \frac{1}{1+ab+b} \geq 1 \\ \\ $Using the condition$ \ \ abc = 1 \\ \\ \frac{1}{1+ ab + b} + \frac{1}{1 + bc +c} + \frac{1}{1+ac+a} = \frac{1}{1+ab+b} + \frac{ab}{ab + b+ 1} + \frac{b}{ab+b+1} \\ \\ =\frac{ab+b+1}{ab+b+1} = 1\\ \\ $Therefore$ \\ \\ \sum_{cyc} \frac{a}{a+b(a^3+1)} \leq \sum_{cyc} \frac{1}{1+b(a+1)} = 1$

$Substituting back in$ \ x,y,z \ $yields the required result$

seanieg89 · Jan 6, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
Great question! Took me a while to get it, but I like my solution:

$\\ $Substitution$ \\ \\ a = \frac{x}{y} , b = \frac{y}{z} , c = \frac{z}{x} \\ \\ \therefore \ abc = 1 \\ \\ \sum_{cyc} \frac{xyz}{x^3 + y^3 + xyz} = \sum_{cyc} \frac{1}{1 + \frac{(x/y)^3 +1}{\frac{xz}{y^2}}} = \sum_{cyc} \frac{a}{a + b(a^3 + 1)} \\ \\ $Now consider$ \\ \\ a^2 + 1 \geq 2a \ \Rightarrow \ \ a^3 + 1 \geq a(a+1) \\ \\ \therefore \ \ \frac{a}{a + b(a^3 +1)} \leq \frac{1}{1 + b(a+1)} \\ \\ $Hence we must prove$ \ \ \ \ \sum_{cyc} \frac{1}{1+ab+b} \geq 1 \\ \\ $Using the condition$ \ \ abc = 1 \\ \\ \frac{1}{1+ ab + b} + \frac{1}{1 + bc +c} + \frac{1}{1+ac+a} = \frac{1}{1+ab+b} + \frac{ab}{ab + b+ 1} + \frac{b}{ab+b+1} \\ \\ =\frac{ab+b+1}{ab+b+1} = 1\\ \\ $Therefore$ \\ \\ \sum_{cyc} \frac{a}{a+b(a^3+1)} \leq \sum_{cyc} \frac{1}{1+b(a+1)} = 1$

$Substituting back in$ \ x,y,z \ $yields the required result$

Very slick solution, perhaps one of your best!

RealiseNothing · Jan 6, 2014

Re: HSC 2014 4U Marathon

seanieg89 said:
Very slick solution, perhaps one of your best!

Can you see what is wrong with mine? I'm fairly sure there is a problem somewhere but I can't find it.

AAEldar · Jan 6, 2014

Re: HSC 2014 4U Marathon

RealiseNothing said:
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 · Jan 6, 2014

Re: HSC 2014 4U Marathon

AAEldar said:
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 · Jan 6, 2014

Re: HSC 2014 4U Marathon

AHHH I GOT IT.

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

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