HSC 2013 MX2 Marathon (archive) (5 Viewers)

RealiseNothing · Jan 6, 2014

Re: HSC 2014 4U Marathon

$x, y, z$ are positive reals such that $\frac{x+y+z}{3} = 1$

Prove that:

$\frac{1}{x^2}(yz-x^3) + \frac{1}{y^2}(xz-y^3) + \frac{1}{z^2}(xy-z^3) \geq 0$

Sy123 · Jan 6, 2014

Re: HSC 2014 4U Marathon

RealiseNothing said:
$x, y, z$ are positive reals such that $\frac{x+y+z}{3} = 1$

Prove that:

$\frac{1}{x^2}(yz-x^3) + \frac{1}{y^2}(xz-y^3) + \frac{1}{z^2}(xy-z^3) \geq 0$

$\\ $Expand$ \ \ $we now must prove$ \\ \\ \frac{xz}{y^2} + \frac{xy}{z^2} + \frac{yz}{x^2} \geq 3$

$\\ $This is immediate through the AM-GM inequality$ \ \ a+b+c \geq 3\sqrt[3]{abc}$

RealiseNothing · Jan 6, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Expand$ \ \ $we now must prove$ \\ \\ \frac{xz}{y^2} + \frac{xy}{z^2} + \frac{yz}{x^2} \geq 3$

$\\ $This is immediate through the AM-GM inequality$ \ \ a+b+c \geq 3\sqrt[3]{abc}$

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.

Trebla · Jan 6, 2014

Re: HSC 2014 4U Marathon

Hate to spoil the party, but you guys reckon you could be a bit more inclusive of the 2014ers? At the moment, it's mainly post-HSC people that are driving this thread.

RealiseNothing · Jan 6, 2014

Re: HSC 2014 4U Marathon

Trebla said:
Hate to spoil the party, but you guys reckon you could be a bit more inclusive of the 2014ers? At the moment, it's mainly post-HSC people that are driving this thread.

Prove that:

$\frac{a+b}{2} \geq \sqrt{ab}$

2014'ers only.

Bonus points: Use circle geometry.

seanieg89 · Jan 6, 2014

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.

RealiseNothing · Jan 6, 2014

Re: HSC 2014 4U Marathon

seanieg89 said:
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 IMO from the 60's and 70's etc are quite doable.

seanieg89 · Jan 7, 2014

Re: HSC 2014 4U Marathon

RealiseNothing said:
Some IMO from the 60's and 70's etc are quite doable.

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.

asianese · Jan 7, 2014

Re: HSC 2014 4U Marathon

Trebla said:
Hate to spoil the party, but you guys reckon you could be a bit more inclusive of the 2014ers? At the moment, it's mainly post-HSC people that are driving this thread.

Entirely?

RealiseNothing · Jan 7, 2014

Re: HSC 2014 4U Marathon

seanieg89 said:
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.

Ah right. Also do you know why there was a massive shift around then?

seanieg89 · Jan 7, 2014

Re: HSC 2014 4U Marathon

RealiseNothing said:
Ah right. Also do you know why there was a massive shift around then?

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 · Jan 7, 2014

Re: HSC 2014 4U Marathon

RealiseNothing said:
Prove that:

$\frac{a+b}{2} \geq \sqrt{ab}$

2014'ers only.

Bonus points: Use circle geometry.

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 for any positive reals $x, y, z$ that:

$\frac{(2x+y+z)^2}{2x^2 + (y+z)^2} + \frac{(2y+x+z)^2}{2y^2+(x+z)^2} + \frac{(2z+x+y)^2}{2z^2+(x+y)^2} \leq 8$

Sy123 · Jan 7, 2014

Re: HSC 2014 4U Marathon

Trebla said:
Hate to spoil the party, but you guys reckon you could be a bit more inclusive of the 2014ers? At the moment, it's mainly post-HSC people that are driving this thread.

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.

$\\ $Find the roots$ \\ \\ S(y) = Ay^3 + By^2 + By + A$

dunjaaa · Jan 7, 2014

Re: HSC 2014 4U Marathon

Here's what I did

Screen shot 2014-01-07 at 1.58.31 AM.png

and also assuming a,b>=0

dunjaaa · Jan 7, 2014

Re: HSC 2014 4U Marathon

are A and B real?

dunjaaa · Jan 7, 2014

Re: HSC 2014 4U Marathon

How do I evaluate this without l'hopital's?

Screen shot 2014-01-07 at 2.02.01 AM.png

Sy123 · Jan 7, 2014

Re: HSC 2014 4U Marathon

RealiseNothing said:
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 for any positive reals $x, y, z$ that:

$\frac{(2x+y+z)^2}{2x^2 + (y+z)^2} + \frac{(2y+x+z)^2}{2y^2+(x+z)^2} + \frac{(2z+x+y)^2}{2z^2+(x+y)^2} \leq 8$

I think I got it

$\\ \sum_{cyc} \frac{(2x+y+z)^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$

RealiseNothing · Jan 7, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
I think I got it

$\\ \sum_{cyc} \frac{(2x+y+z)^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$

gg

that is all

Sy123 · Jan 7, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
How do I evaluate this without l'hopital's? View attachment 29606

One way I've found to do this is to consider the differentiation by first principles of $f(x) = e^x$

$f'(a) = \lim_{h \to 0} \frac{f(a+h) -f(a)}{h}$

$\\ $let$ \ \ f(x) = 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

$\\ e \ $is the number such that$ \ \ \frac{d}{dx}e^x = e^x$

If we define e to be so, then the differentiation by first principles is applicable.

However the more common definition is that

$e= \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n$

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.

$e = \lim_{n \to \infty} (1+1/n)^n$

We use a substitution so we can combine the limit of h to the limit of n, that is; $n =\frac{1}{h}$

$\therefore \ e = \lim_{h \to 0} (1+h)^{1/h}$

$\lim_{h \to 0} \frac{e^h-1}{h} = \lim_{h \to 0} \frac{(1+h)^(1/h \times h) - 1}{h} = 1$

Sy123 · Jan 7, 2014

Re: HSC 2014 4U Marathon

RealiseNothing said:
gg

that is all

haha substitutions can be quite powerful

HSC 2013 MX2 Marathon (archive) (5 Viewers)

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