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I've been thinking to prove each one individually then sum them up which seems too complex to because I haven't made any notable success proving $\frac{x^2}{(x-1)^2}> 1$.

I know that $(x-1)^2\ge 0$ , but I'm not able to transfer it to what I'm looking to prove.

Is there is any techniques to look for when solving these kind of inequalities because I'm totally stuck here.

Michael Rozenberg
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2 Answers2

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Let $x=\frac{a}{b}$, $y=\frac{b}{c}$ and $z=\frac{c}{a}$

Hence,$$\sum_{cyc}\frac{x^2}{(x-1)^2}-1=\frac{(a^2b+b^2c+c^2a-3abc)^2}{(a-b)^2(a-c)^2(b-c)^2}\geq0.$$

Michael Rozenberg
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If $x=1+a, y=1+b, z=1+c$ then $a, b, c \ge 0$ since all are positive integers and

$\begin{array}\\ \dfrac{x^2}{(x-1)^2} + \dfrac{y^2}{(y-1)^2} + \dfrac{z^2}{(z-1)^2} &=\dfrac{(1+a)^2}{a^2} + \dfrac{(1+b)^2}{b^2} + \dfrac{(1+c)^2}{c^2}\\ &=1+\dfrac{1+2a}{a^2} +1+ \dfrac{1+2b}{b^2} +1+ \dfrac{1+2c}{c^2}\\ &=3+\dfrac{1}{a^2}+ \dfrac{1}{b^2}+ \dfrac{1}{c^2}+2(\dfrac{1}{a}+ \dfrac{1}{b} + \dfrac{1}{c})\\ &\ge 3\\ \end{array} $

marty cohen
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  • $xyz=1$ does not rule out the chance that $x<0$ and so $a<0$. – Jack D'Aurizio Jan 05 '18 at 00:43
  • "since all are positive integers" Um... what???? Where on earth did you get that idea. Unless they are all equal to 1, xyz = 1 most certainly means they are not positive integers. – fleablood Jan 05 '18 at 01:08
  • Not only does xyz = 1 not rule out that $x < 0$ unless x,y,z are all equal it assures some of the a,b,c must be negative. – fleablood Jan 05 '18 at 01:12
  • @fleablood Check the edit history, the original question asked, by mistake, for integer solutions. – P. Koymans Jan 05 '18 at 01:12
  • Um, No...https://math.stackexchange.com/posts/2592422/revisions nowhere in the original post were integers asked for which would have obviously have been impossible. unless $|x|=|y|=|z|=1$. – fleablood Jan 05 '18 at 01:15
  • @fleablood It quite clearly says positive integers in the very first post.... – P. Koymans Jan 05 '18 at 01:17
  • @fleablood Quoting the OP here: prove this inequality hold if x,y,z is positive integer. – dxiv Jan 05 '18 at 03:31