Yep there's actually a 1960's IMO inequality that can be done in a few lines with a nice substitution.
will do this tomorrow, but I think it's related to arithmetic and geometric mean (?), and with circle geo, is there something about intersecting secants and squares of a tangent (for RHS at least, not sure what's happening on LHS)?Prove that:
2014'ers only.
Bonus points: Use circle geometry.
You can try to do this variant:
It's called the AM-GM inequality, which is arithmetic mean - geometric mean. The LHS is the AM whilst the RHS is the GM. You are right it is a part of harder 3u, it is the first inequality you will learn and for MX2 definitely the most important.
Yep I have seen the proof, but I can't recall the specifics
I did it just then without circle geo, but not 100% sure if I did it right as a "prove" q, no idea how to do with circle geo though.
I think there are some subtleties with the last line about double limits you need to clarify..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;
^(1/h \times h) - 1}{h} = 1)
Can you give examples of questions you posted which were 2014er friendly? I think the amount of higher level Maths being demonstrated between you guys is intimidating to the 2014ers who are learning things for the first time, which influences whether they post or not.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.
 = Ay^3 + By^2 + By + A )
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).I think I got it
^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 )
Alternatively:So Jensen's doesn't work, but I found a solution:
\\ \leq \frac{1}{3}\sum_{\textrm{cyc}}(1+3(4x+1))=4+4(x+y+z)=8.$ $)
How is that a solution though? That is just noting that the LHS expression is homogenous, which is what allowed me to make the assumption the variables have sum 1. Sy did the same thing explicitly, by dividing by x+y+z.Alternatively:
Letand
and
and the inequality still holds.
So it suffices to prove the inequality for some value of C where
The easiest value of C to use is 3, so letand it's fairly easy from there.
Letting
Lol, would hardly call that an alternate solutionLetting
by polynomial division
 = \frac{4}{3} (6) = 8)
. It's like exactly the same difficulty.
Well yeah, that's what I did above. Really doesn't change the time it takes to do it.Though I guess it would work withbut you would need to multiply/divide by a factor of 3 to get the polynomial division done easily.