RealiseNothing
what is that?It is Cowpea
Re: HSC 2014 4U Marathon
are positive reals such that 
Prove that:
 + \frac{1}{y^2}(xz-y^3) + \frac{1}{z^2}(xy-z^3) \geq 0)
Prove that:
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HSC 2013 MX2 Marathon (archive) (6 Viewers)
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Sy123
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RealiseNothing
what is that?It is Cowpea
Re: HSC 2014 4U Marathon
Yep that was a bit on the easier side I guess. It's hard trying to think of a really challenging inequality, I'm trying though lol.
RealiseNothing
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seanieg89
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Re: HSC 2014 4U Marathon
Some national olympiad problems from Copy of Kalva homepage are the perfect level for this thread. Check that any given problem can be done using only mx2 stuff though. And probably avoid IMO stuff outside of Q1,4.
Some national olympiad problems from Copy of Kalva homepage are the perfect level for this thread. Check that any given problem can be done using only mx2 stuff though. And probably avoid IMO stuff outside of Q1,4.
RealiseNothing
what is that?It is Cowpea
seanieg89
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Re: HSC 2014 4U Marathon
Oh definitely. There is a huge shift in difficulty somewhere between 75 and 95. (I probably still can't do most modern Q6s, and certainly not in the exam conditions.) I meant any inequality that is not Q1 or 4 in relatively recent olympiads.
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RealiseNothing
what is that?It is Cowpea
Re: HSC 2014 4U Marathon
Ah right. Also do you know why there was a massive shift around then?
seanieg89
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Re: HSC 2014 4U Marathon
Beats me. Maybe the fact that countries were starting to prepare more intensively meant the organising committee thought it needed to be made harder. There is probably an explanation online somewhere but am too lazy to look.
RealiseNothing
what is that?It is Cowpea
Re: HSC 2014 4U Marathon
Prove for any positive reals that:
^2}{2x^2 + (y+z)^2} + \frac{(2y+x+z)^2}{2y^2+(x+z)^2} + \frac{(2z+x+y)^2}{2z^2+(x+y)^2} \leq 8)
I think we should post 2014'er only questions and questions in general for anyone. The above is my 2014'er question. Here is a question from seanie's link for everyone:Prove that:
2014'ers only.
Bonus points: Use circle geometry.
Prove for any positive reals
Last edited:
Sy123
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Re: HSC 2014 4U Marathon
 = Ay^3 + By^2 + By + A )
Problem is when I do try and engage the 2014er friendly topics no one answers my questions, nor are they trying to post questions themselves.
Sy123
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Re: HSC 2014 4U Marathon
^2}{2x^2+(y+z)^2} = \sum_{cyc} \frac{\left(1 + \frac{x}{x+y+z} \right)^2}{2 \left(\frac{x}{x+y+z} \right)^2 + \left(\frac{y+z}{x+y+z} \right)^2} \\ \\ $Substitute$ \ \ \ a = \frac{x}{x+y+z} , \ b= \frac{y}{x+y+z} , \ c = \frac{z}{x+y+z} \\ \\ \therefore \ \ a+b+c = 1 \\ \\ \therefore \ \ \sum_{cyc} \frac{(2x+y+z)^2}{2x^2+(y+z)^2} = \sum_{cyc} \frac{(a+1)^2}{2a^2 + (b+c)^2} = \sum_{cyc} \frac{a^2 + 2a +1}{3a^2 - 2a + 1} \\ \\ $Take Jensen's inequality which is reversed for concave functions where$ \\ \\ f(x) = \frac{x^2+2x+1}{3x^2-2x+1} \ \ $then$ \\ \\ \frac{1}{3} (f(a) + f(b) + f(c) ) \leq f(\frac{1}{3}(a+b+c)) = f \left(\frac{1}{3} \right) \\ \\ $Therefore$ \\ \\ \sum_{cyc} \frac{(2x+y+z)^2}{2x^2+(y+z)^2} = \sum_{cyc} \frac{a^2 + 2a +1}{3a^2 - 2a + 1} \leq f \left( \frac{1}{3} \right) = 8 \\ \\ \square )
I think I got itI think we should post 2014'er only questions and questions in general for anyone. The above is my 2014'er question. Here is a question from seanie's link for everyone:
Prove for any positive reals
RealiseNothing
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Sy123
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Re: HSC 2014 4U Marathon
 = e^x )
 = \lim_{h \to 0} \frac{f(a+h) -f(a)}{h} )
 = e^x \ \ a = 0 \\ \\ e^0 = \lim_{h \to 0} \frac{e^h - 1}{h} = 1 )
However some things need to be noted here, in how we define the number e.
If we define the number e to be
If we define e to be so, then the differentiation by first principles is applicable.
However the more common definition is that
^n )
If we accept this definition, then using differentiation by first principles in the way that I did, would be circular in nature. Because we need that limit of e, in order to find that d/dx e^x = e^x (not sure if I explained correctly).
So using the second definition.
^n )
We use a substitution so we can combine the limit of h to the limit of n, that is;
^{1/h} )
^(1/h \times h) - 1}{h} = 1)
One way I've found to do this is to consider the differentiation by first principles of
However some things need to be noted here, in how we define the number e.
If we define the number e to be
If we define e to be so, then the differentiation by first principles is applicable.
However the more common definition is that
If we accept this definition, then using differentiation by first principles in the way that I did, would be circular in nature. Because we need that limit of e, in order to find that d/dx e^x = e^x (not sure if I explained correctly).
So using the second definition.
We use a substitution so we can combine the limit of h to the limit of n, that is;
Sy123
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