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Re: HSC 2014 4U Marathon Well done, I don't know exactly what you did but the general idea was just doing double angle rule (which you did)

Re: HSC 2014 4U Marathon Now try this \\ $find$ \ \ \ \cos^2 \theta + \cos^2 2\theta + \dots + \cos^2 n \theta

Re: HSC 2014 4U Marathon Yea something like that

Re: HSC 2014 4U Marathon \\ \sqrt[n]{m+1} < \frac{(1+1+...+1) + (m+1)}{n} = \frac{n-1 + m+1}{n} = \frac{m+n}{n} \ \ \ \ \ \fbox{1} \\ \\ \sqrt[m]{m+1} < \frac{m-1 + n+1}{m} = \frac{n+m}{m} \ \ \ \ \ \fbox{2} \\ \\ \frac{1}{\fbox{1}} + \frac{1}{\fbox{2}} \Rightarrow \ \ \frac{1}{\sqrt[n]{1+m}}...

Re: HSC 2014 4U Marathon I recall some: \\ $Prove that the angle in a semicircle is a right angle using complex numbers$ \\ $The 4 roots of unity and 3 roots of unity are plotted on the argand diagram, and form a square and triangle respectively, find the area of the overlapping area$

Re: HSC 2014 4U Marathon Yep I have seen the proof, but I can't recall the specifics

Re: HSC 2014 4U Marathon You can try to do this variant: a + \frac{1}{a} \geq 2 via calculus if you want, but I don't think you can use calculus for a 2 variable one

Re: HSC 2014 4U Marathon haha substitutions can be quite powerful

Re: HSC 2014 4U Marathon 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...

Re: HSC 2014 4U Marathon 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 =...

Re: HSC 2014 4U Marathon 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

Re: HSC 2014 4U Marathon \\ $Expand$ \ \ $we now must prove$ \\ \\ \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}

Re: HSC 2014 4U Marathon 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...

Re: MX2 Integration Marathon \int_0^1 \frac{\ln x}{\sqrt{1-x^2}} \ dx

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

Re: HSC 2014 4U Marathon Aha ok then Demonstration?

Re: HSC 2014 4U Marathon \\ $Help needed$ \\ \\ $Prove$ \\ \\ \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...

Re: HSC 2014 4U Marathon Here is my long solution; \\ (\sqrt{a_1} + \sqrt{a_2} + \dots + \sqrt{a_n})^2 = (a_1 + \dots + a_n) + 2(\sqrt{a_1a_2} + \dots) \\ \\ $Use$ \ a_k + a_m \geq 2\sqrt{a_k a_m} \\ \\ \therefore \ 1 + (n-1)(a_1 + \dots + a_n) \geq (\sqrt{a_1}+ \dots + \sqrt{a_n})^2 \\ \\...

Re: HSC 2014 4U Marathon Here is my attempt: \sum_{k=1}^{\infty} \sin \left(\frac{1}{k} \right) = \sum_{k=1}^{\infty} \cos \left(\frac{\pi}{2} - \frac{1}{k} \right) \\ $Consider the graph$ \ y= \cos x \ \ \ \ 0 \leq x \leq \frac{\pi}{2} \ $and the straight line through$ \ (0,1) \...