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Re: HSC 2014 4U Marathon So Jensen's doesn't work, but I found a solution: $Calculus shows that $3x^2-2x+1\geq 2/3$ for all $x$.\\ \\ So: \\ \\ $S=\sum_{\textrm{cyc}} \frac{x^2+2x+1}{3x^2-2x+1}=\frac{1}{3}\sum_{\textrm{cyc}}\left(1+\frac{8x+2}{3x^2-2x+1}\right)\\ \leq...

Re: HSC 2014 4U Marathon That was my first attempt, but alas: second derivative of (x^2+2x+1)/(3x^2-2x+1) - Wolfram|Alpha, we do not have fixed concavity in (0,1). Our best friend Jensen is of no help here unless we split into cases or something, but that looks like a pretty ugly idea.

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.

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.

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.

Re: HSC 2014 4U Marathon Very slick solution, perhaps one of your best!

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.

Re: HSC 2014 4U Marathon good :).

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.

Some of the things I got marked wrong for in high school were just ridiculous. Eg1. A question showed a graph and asked you to provide a possible function which had this graph. Apparently my |1-x| was incorrect, because the marker was looking for |x-1|. Eg2. A marker did not understand my...

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...

Re: HSC 2014 4U Marathon Uh, no they aren't.

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.

Re: HSC 2014 4U Marathon If you take the log of both sides, it is a straightforward application of Jensen's inequality.

All of them really.

Re: HSC 2014 4U Marathon Nice! Yeah imo Jensen's is extremely powerful for how easy it is to prove.

It would be cool if people could volunteer to upload their papers on a thread here, for the benefit of future students to see how many marks various partial solutions and mistakes are worth.

Looks good to me (apart from changing the transformed variable from t to p from the first line to the third line of calculating the Laplace transform). You can even use this as a way of defining fractional differentiation. It is a common result, we often get these kind of interactions between...

Re: HSC 2014 4U Marathon

Re: HSC 2014 4U Marathon Haha the gap in lengths would probably be at least partially closed by me having to prove Jensen's before using it. I will post a statement and proof of it here later today, just out of interest to see how compact and elementary I can make it.