AHZOLFAGHARI wrote: $x,y,z$ are three real number not equal to zero such that $x+y+z=xyz$ Prove that $$ \sum (\frac{x^2-1}{x})^2 $$ is equal or more than 4 After homogenization we need to prove that $\sum_{cyc}(x^2+xy+xz-yz)^2\geq4xyz(x+y+z)$, which is a fourth degree symmetrical inequality of three variables. Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative and $xyz=w^3$. Hence, we need to prove that $f(w^3)\geq0$, where $f$ is a linear function. Thus, $f$ gets a minimal value for an extremal value of $w^3$, which happens for an equality case of two variables. Since the last inequality is homogeneous of even degree, we can assume $y=z=1$, which gives $(x-1)^2(x+3)^2\geq0$. Done!

I have got a solution. The given expression on the LHS $\equiv$ $(x-\frac{1}{x})^2+(y-\frac{1}{y})^2+(z-\frac{1}{z})^2$.Hence, $\Leftrightarrow$ $x^2+y^2+z^2+(\frac{1}{x})^2+(\frac{1}{y})^2+(\frac{1}{z})^2 \ge 10$ Now, $$x^2+y^2+z^2 \ge 9$$by AM-GMso we have to prove that $(\frac{1}{x})^2+(\frac{1}{y})^2+(\frac{1}{z})^2 \ge 1$ Amazingly we have from the expression $x+y+z=xyz$ $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1$ which directly gives us $(\frac{1}{x})^2+(\frac{1}{y})^2+(\frac{1}{z})^2 \ge (\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx})$ which is trivial

I am poor at LATEX .I just learnt it today

arqady wrote: $\sum_{cyc}(x^2+xy+xz-yz)^2\geq4xyz(x+y+z)$ $\sum_{cyc}(x^2+xy+xz-yz)^2-4xyz(x+y+z)=\frac{1}{2}\sum_{cyc}{(x-y)^2(x+y+2z)^2}\geq0$

SohamSchwarz119 wrote: $$x^2+y^2+z^2 \ge 9$$ It's wrong. Try $x=1$, $y=z=1-\sqrt{2}$.

sorry, I did not notice that they are not positive

This problem was proposed by Amin Fathpour .

Since $x + y + z = xyz$, it is well-known that there exists a triangle $\triangle ABC$ for which $x = \tan A, \; y = \tan B, \; z = \tan C.$ Then from the identity $\tan 2\theta = \tfrac{2\tan \theta}{1 - \tan^2 \theta}$, we find that \[\sum\limits_{\text{cyc}} \left(\frac{x^2 - 1}{x}\right)^2 = \sum\limits_{\text{cyc}} \left(\frac{\tan^2 A - 1}{\tan A}\right)^2 = 4\sum\limits_{\text{cyc}} \cot^2 2A.\] Now let $f(x) := \cot^2 x.$ A straightforward derivative computation shows that $f''(x) = 2\csc^4 x(\cos 2x + 2) \ge 0.$ Thus by Jensen's Inequality we have \[\sum\limits_{\text{cyc}} \cot^2 A \ge 3\cot^2\left(\frac{2A + 2B + 2C}{3}\right) = 3\cot^2\left(120^{\circ}\right) = 1. \; \square\]

From $ \sum x = xyz $ we can conclude that : \[ x(1-y^2)(1-z^2) + y(1-z^2)(1-x^2) + z(1-x^2)(1-y^2) = 4xyz \] Now we have : \[ \sum ( \frac {x^2-1}{x})^2 \ge \sum ( \frac {y^2-1}{y} )( \frac {z^2-1}{z} ) \] \[ = \frac { \sum x(y^2-1)(z^2-1) }{xyz} = 4 \]

$x+y+z=xyz\implies xyz\geq3\sqrt{3}$. $(\frac{x^2-1}{x})^2 =\frac{8x^2}{9}+\frac{x^2}{9}+\frac{1}{x^2}-2\geq\frac{8x^2}{9}-\frac{4}{3}$, $ (\frac{x^2-1}{x})^2+(\frac{y^2-1}{y})^2 + (\frac{z^2-1}{z})^2 \geq\frac{8}{9}( x^2+y^2+z^2)-4\geq\frac{8}{3}\sqrt[3]{(xyz)^2}-4\geq4 .$

War-Hammer wrote: sqing wrote: $$x+y+z=xyz\implies xyz\geq3\sqrt{3}$$ Excuse me , but the numbers aren't necessary positive.

Use the fact that in any triangle (so with $A+B+C=\pi$) we have $ tan(A)+tan(B)+tan(C)=tan(A)tan(B)tan(C) $

GoJensenOrGoHome wrote: Use the fact that in any triangle (so with $A+B+C=\pi$) we have $ tan(A)+tan(B)+tan(C)=tan(A)tan(B)tan(C) $ $A\neq\frac{\pi}{2},B\neq\frac{\pi}{2},C\neq\frac{\pi}{2}.$

AHZOLFAGHARI wrote: $x,y,z$ are three real number not equal to zero such that $x+y+z=xyz$ Prove that $$ \sum (\frac{x^2-1}{x})^2 $$ is equal or more than 4 propossed by Amin Fathpour Aren't $x,\ y,\ z$ be restricted to positive real numbers?

AHZOLFAGHARI wrote: $x,y,z$ are three real number not equal to zero such that $x+y+z=xyz$ Prove that $$ \sum (\frac{x^2-1}{x})^2 $$ is equal or more than 4 propossed by Amin Fathpour The problem is very easy for Iranian Mathematical Olympiad. If $x,\ y,\ z$ were positive real numbers, then the proof is as follows. Using the AM-GM inequality, we have $x+y+z=xyz\geq 3\sqrt[3]{xyz}$, yielding $\sqrt[3]{(xyz)^2}\geq 3$, thus $\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2+\left(z-\frac{1}{z}\right)^2$ $=x^2+y^2+z^2+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-6$ $\geq 3\sqrt[3]{x^2y^2z^2}+\frac{3}{\sqrt[3]{x^2y^2z^2}}-6$ $\geq 3\left(3+\frac 13\right)-6=4$ Because the function $y=t+\frac{1}{t}\ (t>0)$ decreases monotonous on $0< t\leq 1$ and increase monotonous on $t\geq 1.$

My solution: $ \sum (\frac{x^2-1}{x})^2 =\frac{\sum((x^2-1)yz)^2}{(xyz)^2}$ which now gives us equivalent to $\sum((x^2-1)yz)^2\ge 4x^2y^2z^2$ equivalent to $\sum x^4y^2z^2+\sum x^2y^2\ge 10x^2y^2z^2$ which is now for the first sum $\sum x^4y^2z^2=x^2y^2z^2(\sum x^2)\ge 9x^2y^2z^2$(by condition $\sum x=xyz\le (\frac{\sum x}{3})^3 \implies (\sum x)^2\ge 27 \implies 3(\sum x^2)\ge (\sum x)^2 \ge 27 \implies (\sum x^2) \ge 9$). which is now equivalent to $\sum x^2y^2\ge x^2y^2z^2=(\sum x)^2 $ so we multiply by $x+y+z=xyz$ and we get that its equvialent to $ \sum (x^3y^2+y^3x^2)+\sum x^2y^2z \ge \sum x^3yz+2\sum x^2y^2z$ which is $AMGM$(or Muirhead we have $3,2,0$ and $3,1,1$ and$2,2,0$) .

my solution = let $x=tan A , y=tan B$ and $z=tan C$ and by given condition $A,B,C$ are angles of a triangle. than we have $ \sum (\frac{x^2-1}{x})^2 =\sum (tan A-cotA)^2=\sum tanA^{2} +\sum cotA^{2}-6$ also $\sum tanA^{2} + \sum cotA^{2} = \sum 4(cosec2A)^2 - 6$ so it remains to prove that $\sum (cosec2A)^2=3+\sum (cot 2A)^2 \ge 4$ which is $\sum (cot 2A)^2\ge 1$ now as $cotx$ is concave function hence by jensen inequality we have $\sum (cot 2A)^2\ge 3(cot 120)^2 = 1$ so we are done !!!

We cannot use Jensen's inequality directly: f(x) is not defined at some points in the whole interval. We must divide in two cases.

AHZOLFAGHARI wrote: $x,y,z$ are three real numbers inequal to zero satisfying $x+y+z=xyz$. Prove that $$ \sum (\frac{x^2-1}{x})^2 \geq 4$$ Proposed by Amin Fathpour We have \[LHS-RHS=\frac{1}{2}\sum{\frac{(xy+1)^2(x-y)^2}{x^2y^2}}\ge{0}\]

Does anyone have a solution by Jensen? (I mean not by trigonometry!)

Is this problem possibly solved by T2?

K.N wrote: Does anyone have a solution by Jensen? (I mean not by trigonometry!) Here you go. Let $l=x+y+z$, it is $s = xyz \le (\dfrac{x+y+z}{3})^3=s^3/27$ $\implies$ $s \ge 3\sqrt{3}$. Note tho that the function $f(x)=(\dfrac{x^2-1}{x})^2$ is convex. Hence, applying Jensen , $$ \sum (\frac{x^2-1}{x})^2 \geq 3 (\frac{(s/3)^2-1}{s/3})^2 $$, which leads to the result.

Let $(x,y,z)=(\tan A, \tan B, \tan C)$ where $A,B,C$ are angles of a triangle. Then the sum is equal to \[\sum_{cyc} \left(\frac{\tan^2 A -1}{\tan A}\right)^2=4\sum_{cyc} {\cot^2 2A}\]and since $f(x)=\cot^2 2x$ has its 2nd derivative equal to $f''(x)=8\csc^2 2x(\csc^2 2x+2\cot^2 2x)\ge 0$ , we have \[\sum_{cyc} \left(\frac{\tan^2 A -1}{\tan A}\right)^2=4\sum_{cyc} {\cot^2 2A} \ge 4 \sum_{cyc} 3\left(\cot^2 \left(\frac{2A+2B+2C}{3}\right)\right)=12\cot^2 (120^{\circ})=4. \blacksquare\] @Below: Oops, I didn't realise it at the first place, trigonometric sub would work.

Keith50. Actually, our variables can be negatives.

$\forall a,b,c \in \mathbb{R}$, we have the inequality $a^2+b^2+c^2 \geq ab+bc+ca$. From the given condition, we have $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1$. Also, $$\sum_{cyc} \left(x - \frac{1}{x}\right)^2 \geq \sum_{cyc} \left(x-\frac{1}{x}\right)\cdot \left(y-\frac{1}{y}\right) =1+ \sum_{cyc} xy \ -\sum_{cyc}\frac{x+y}{z}$$Which, indeed, equals 4 because $x+y= z\cdot(xy-1)$ etc. Hence the desired inequality holds.