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If 0<z\leq y\leq x \ \textrm{prove} \frac{x^{2}y}{z} + \frac{y^{2}z}{x} + \frac{z^{2}x}{y} \geq x^{2} + y^{2} + z^{2}

if a,b,c,d,e is some order of numbers from 1 to 5. What is the maximum possible value of S=ab+bc+cd+de+ea

\textrm{Find the minimum number of points on a complex plane such that} \textrm{for any arbitrary point on the plane at least one the distance between that arbitrary point} \textrm{to the chosen points is irrational.} \textrm{For instance, suppose the minimum number of points} \textrm{...

\textrm{Solve x} (2^{x} -4)^{3} + (4^{x} -2) ^{3} = (4^{x}+ 2^{x} -6)^{3}

\textrm{1.If} \ f(\frac{x-1}{x+1})+f(\frac{-1}{x}) + f(\frac{1+x}{1-x}) = x. \textrm{Find f(x)}

\textrm{1.If X is a set with n distinct elements.} \textrm{then prove the number of pairs (A,B) is } \ 3^{n} - 2^{n} \textrm{where} \ A \subset B \ \textrm{and} \ A,B \subseteq X \textrm{2.We call a 10 digit number interesting if all the digits are different} \textrm{and the number...

Re: HSC 2017 MX2 Marathon \textrm{for} x,y,z \in \mathbb{R}\ xyz(x+y+z) =1 \textrm{Find the minimum of the following expression}\ (x+y)(y+z)(x+z)

Re: HSC 2017 MX2 Integration Marathon \textrm{1.If} \ \alpha, \beta \ \textrm{ are real numbers such that} \beta^{3} -6\beta^{2} + 13\beta = 19 , \ \alpha^{3}-6\alpha^{2} +13\alpha = 1 \textrm{Hence, find } \ (\alpha + \beta) \textrm{2.If }f(x) = \frac{4^{x}}{4^{x}+2} \textrm{Find}...

\textrm{1. Suppose}\ a,b,c \in \mathbb{R} \ \textrm{such that} 9a+11b+29c =0 \ \textrm{prove } \ ax^{3}+bx^{2}+c = 0, \textrm{has a root between } \ 0 \leq x \leq 2 \textrm{2.If polynomial} \ P(x) = ax^{3} + bx+ c \ \textrm{is divisible by} \ G(x) = x^{2} + tx+1 \textrm{Find a...

Re: HSC 2017 MX2 Integration Marathon \int \frac{x^2+2x+1}{x^2-3x-1} dx = \int \frac{x^2-3x-1}{x^2-3x-1} dx +\frac{5}{2} \int \frac{2x-3}{x^2-3x-1} dx + \frac{23}{2} \int \frac{1}{x^2-3x-1} dx = \int \frac{x^2-3x-1}{x^2-3x-1} dx +\frac{5}{2} \int \frac{2x-3}{x^2-3x-1} dx +...

This is what I got: the first inequality: (b-c)^{2} \geq 0 , (a-c)^{2} \geq 0 , (a-b)^{2} \geq 0 adding all of them together we get: 2(a^{2} + b^{2} +c^{2}) \geq 2(ab + ac +bc) (a^{2} + b^{2} +c^{2}) \geq (ab + ac +bc) therefore, (a+b+c)^{2} \geq 3(ab + ac +bc) the second...

1. Suppose the roots of the original equation are \alpha , \beta, \gamma Using sum and products of the roots we know: \alpha + \beta + \gamma = \frac{2}{3}, \alpha \ \beta + \alpha \ \gamma + \beta \ \gamma = 0 , \alpha \ \beta \ \gamma = -\frac{1}{3} To create a polynomial which...

I used induction to solve the question, I'm curious to know whether there is any other way to go about the question: Proof: Base Case: n=2 LHS = \frac{1}{\sqrt{a_{2}}+ \sqrt{a_{1}}} = RHS Suppose it is true for n=k: \frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}} + .... +...

The way I solved it uses the identity: tan2x = \frac{2tanx}{1-tan^{2}x} \textrm{or} \ 1-tan^{2}x = \frac{2tanx}{tan2x} and simplifies the question to: P_{m} = cota \ 2^{m}\ tan(\frac{a}{2^{m}}) then: lim_{m\to\infty}P_{m} = a \ cota

well, the solution I came up with relies on induction, but I believe there is another way to solve it as well. Proof: prove the base case: n=2 LHS = tan(\alpha) \ tan(2\alpha) = tan(\alpha)*\frac{2\ tan(\alpha)}{1-tan^{2}(\alpha)} = \frac{2tan^{2}(\alpha)}{1-tan^{2}(\alpha)} =...

That proves the above equation is a general form of a circle, but the question says that it also is the form of a line in the complex plane. To just add on to your proof: 1.prove the equation below represents a line in complex plane: \alpha z\overline{z} + \beta \ z +\overline{\beta}\...

\textrm{It is one of the definition of product/sum of infinite terms:} \textrm{Let me give you one example of infinite product,} \frac{3}{2}\ \frac{5}{4} \ \frac{7}{6}\ \frac{9}{8}\ .... \textrm{note that every term is greater than 1 so either the product is infinite or a finite number...

You are making the same mistake. instead of infinity you should find S_{k} for finite k that means S_{k} = \prod_{n=2}^{k} (\frac{n-1}{n+1}) \ \prod_{n=2}^{k}(\frac{2n+1}{2n-1}) = (\frac{2}{k(k+1)})(\frac{2k+1}{3}) = \frac{4k+2}{3k^{2}+3k} then lim_{k\to\infty} S_{k} = lim_{k\to\infty}...

well, I used your method, except I chose to work with base a. But I like challenge and as you suggested, there might be a faster method. I tried to find a different or possibly faster method, and this is what I got: log_{a}\ 16 +log_{\sqrt{2}}\ a = 9 \ \ \ \ \ \ (1) log_{a}\ 16 - 8 =...

Unfortunately, the answer is not \frac{2}{3} . You have an interesting idea but when it gets to infinite product or sum, weird things happen. I believe, you can't in general use telescoping method for infinite sums or products. Consider this example: let s =...