Let $ a=\frac{x}{1-x},b=\frac{y}{1-y},c=\frac{z}{1-z}$ Then we have : $ abc=(a+1)(b+1)(c+1)=\frac{1}{(x-1)(y-1)(z-1)}$ $ \Leftrightarrow ab+ac+bc+a+b+c+1=0$ Therefore : $ a^2+b^2+c^2=a^2+b^2+c^2+2(ab+ac+bc)+2(a+b+c)+2$ $ \Leftrightarrow a^2+b^2+c^2=(a+b+c+1)^2+1\geq 1$ So problem a claim . The equality hold if and only if $ a+b+c+1=0$ This is equivalent : $ xy+zx+zx=3$ From $ x=\frac{1}{yz}$ we have $ \frac{1}{z}+\frac{1}{y}+yz=3$ $ \Leftrightarrow z^2y^2-y(3z-1)+z=0$ $ \Delta =(3z-1)^2-4z^3=(z-1)^2(1-4z)$ We only chose $ z=\frac{1-m^2}{4},|m|>0$ then the equation has rational solution $ y$ Because $ x=\frac{1}{yz}$ so it also a rational . Problem claim .

What is the point of an IMO problem consisting to 100% of computation? For part (i), you find that (1) $ \frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} - 1 = \frac {\left(yz+zx+xy - 3\right)^2}{\left(x - 1\right)^2\left(y - 1\right)^2\left(z - 1\right)^2}$. [Thanks to Pi3B and droid347 for correcting computation mistakes here.] This is trivial to find using a CAS with a factor command and the substitution $ x = \frac {a}{b}$, $ y = \frac {b}{c}$, $ z = \frac {c}{a}$. Without a CAS, prepare for some annoying work - but it's still quite natural to expect that the numerator of the resulting fraction is a square of some polyomial, and finding that polyonomial is easily reduced to a big and ugly but doable system of quadratic equations. For part (ii), you are looking for rational $x$, $y$, $z$ with $xyz = 1$ and $x + y + z = 3$. In other words, you are looking for rational $ u$, $ v$, $ w$ with $ uvw = 1$ and $ u + v + w = 3$ (indeed, we have just set $u = \frac{1}{x}$, $v = \frac{1}{y}$ and $w = \frac{1}{z}$). In other words, you are looking for rational $u$ and $v$ with $u + v + \frac {1}{uv} = 3$. This rewrites as $ v^2 + \left(u - 3\right)v + \frac {1}{u} = 0$, which is a quadratic equation in $v$. So for a given $u$, it has a rational solution $v$ if and only if its determinant $ \left(u - 3\right)^2 - 4\cdot\frac {1}{u}$ is a square. But $ \left(u - 3\right)^2 - 4\cdot\frac {1}{u} = \frac {u - 4}{u}\left(u - 1\right)^2$, so this is equivalent to $ \frac {u - 4}{u}$ being a square. Parametrize... Except of the formula (1), I fail to see anything interesting in the problem - but even this formula is merely a result of lengthy computations. Is this the future of olympiad problems? darij

Indeed, a terrible problem! However, it worths noticing that it is more elegant than the usual nasty things they give at IMO. It remains however a very bad choice: what I simply did was to replace $ z=1/(xy)$ in the inequality, pass the term in $ z^2/(z-1)^2$ on the RHS and simply compute things with the substitution $ x+y=s, xy=p$. Incredibly, 5 minutes of computations and surprise, it appears in the form $ s^2+2s(p^2-3p)+p(p^3-6p^2+9p)\geq 0$, which is obviously the same as $ (s+p^2-3p)^2\geq 0$. So nothing smart, really nothing. I'm wondering however how the author of this problem found the result-I think that's pretty curious. As for the second question, that's really put there for which reason? To make this a problem 2-type?

Well... Quote: $ \frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} - 1 = \frac {\left(x + y + z - 3\right)^2}{\left(x - 1\right)^2\left(y - 1\right)^2\left(z - 1\right)^2}$ I think there is a little mistake in your expression... Actually, $ (x,y,z) = ( - 2, - 2,\frac {1}{4})$ is an equality case but the RHS of your equality is not zero with these values...

replace $ (x + y + z - 3)^2$ with $ (xy + yz + zx - 3)^2$ with substitution $ x = \frac {1}{a}$ it remains to prove that $ (a + b + c - 3)^2 \geq 0$ so i think that that's the formula he wanted..

Yes. It seems not a very clever choice for IMO problem. I have to say that this problem is very old (my friend discovered it and I did put it in Secrets In Inequalities). The original problem is (which I think it is better in term of aesthetic): \[ \frac{x^2}{(x-y)^2}+\frac{y^2}{(y-z)^2}+\frac{z^2}{(z-x)^2}\ge 1.\] This problem is transformed to \[ \left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}-3\right)^2\ge 3.\] Part (ii) is certainly too easy if one can finish (i).

My solution to part b) is totally the same as TTsphn's one. Solution to part a): Let $ a=1-\frac{1}{x}$ and so on... Then our inequality becomes: $ a^2b^2+b^2c^2+c^2a^2\geq a^2b^2c^2$,while $ (1-a)(1-b)(1-c)=1$. Second condition gives us that: $ a^2b^2+b^2c^2+c^2a^2=a^2b^2c^2+(a+b+c)^2$,that's all,I guess. As it was said problem today was easy enough,and as I know two guys from Ukrainian team and three guys from Russian team solved all the problems and I have a sense that most of Chinese team did all the problem.

Why is it important to have 3 variables and not just $ n$ variables? I mean, isn't the following true? "Let $ a_1, a_2, ..., a_n$ be real numbers such that $ \prod a_i = 1$ and $ a_i \neq 1$, then $ \sum \frac {(a_i)^2}{(a_i - 1)^2} \geq 1$" For $ n=2$ it's obviously true and $ n=3$ is IMO 2008 P2 What about $ n > 3$?

In part b) common answer seems to be following: $ x = - \frac {d}{e} - \frac {d^2}{e^2};$ $ y = - \frac {e}{d} - \frac {e^2}{d^2};$ $ z = \frac {de}{(d + e)^2}.$ $ d,e$ are integers such that $ d,e,d+e \not= 0$.

Combinatorics? It's #6. Well, maybe one of the others- but we have only seen half the test so far. Even by hand, eliminating a variable by setting $ z=\frac1{xy}$ and doing nothing clever, it only took twenty minutes or so to find a version of darij's (1). Something like that was inevitable, given the hint of (ii) that there's a curve of solutions.

for part $ a$ if you try $ x\rightarrow\dfrac{1}{x}$ and expand it easily turns to $ (x + y + z)^2 + 9\geq 6(x + y + z)$... for the part $ b$ you can take $ (x,y,z) = \left( - k(k + 1),\dfrac{ - (k + 1)}{k^2},\dfrac{k}{(k + 1)^2}\right)$

I don't know why this problem has to be blamed.. This is a cool inequality I think. Anyhow here is my solution, which seems to be intuitive. First letting $ x=\frac{q}{p}, y=\frac{r}{q}, z=\frac{p}{r}$, we have to show that $ \sum \frac{q^2}{(p-q)^2} \geq 1$ holds. Define $ f(t)$ to be $ \sum \frac{(t+q)^2}{(p-q)^2} = (\sum \frac{1}{(p-q)^2}) t^2 + 2(\sum \frac{q}{(p-q)^2})t + \sum \frac{q^2}{(p-q)^2} = At^2+2Bt+C$. This is a quadratic function of $ t$ and we know that this has minimum at $ t_0$ such that $ At_0 + B = 0$. Hence, $ f(t) \geq f(t_0) = At_0^2+2Bt_0+C = Bt_0+C = \frac{AC- B^2}{A}$. Since $ AC-B^2 = (\sum \frac{1}{(p-q)^2})(\sum \frac{q^2}{(p-q)^2}) - (\sum \frac{q}{(p-q)^2})^2$ and we have $ (a^2+b^2+c^2)(d^2+e^2+f^2)-(ad+be+cf)^2 = \sum (ae-bd)^2$, we obtain $ AC-B^2 = \sum (\frac{r-q}{(p-q)(q-r)})^2 = \sum \frac{1}{(p-q)^2}=A$. This makes $ f(t) \geq 1$, as desired. The second part is trivial, since we can find $ (p,q,r)$ with fixed $ p-q$ and any various $ q-r$, which would give different $ (x,y,z)$ satisfying the equality.

delegat wrote: If $ x$, $ y$ and $ z$ are three real numbers, all different from $ 1$, such that $ xyz = 1$, then prove that $ \frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} \geq 1$. $ \frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} - 1$ $ \equiv\frac {a^{6}}{\left(a^3 - abc\right)^{2}} + \frac {b^{6}}{\left(b^3 - abc\right)^{2}} + \frac {c^{6}}{\left(c^3 - abc\right)^{2}} - 1$ $ = \frac {(bc + ca + ab)^2\left(b^2c^2 + c^2a^2 + a^2b^2 - a^2bc - b^2ca - c^2ab\right)^2}{\left(a^2 - bc\right)^2\left(b^2 - ca\right)^2\left(c^2 - ab\right)^2}\geq 0.$

The inequality this year is too bad, the problem is not new. You can see it is equivalent to the following in my collection "nothing1.pdf" (i have posted it on Mathlinks) \[ \sum \frac{a^2}{(a-b)^2} \ge 1\] And my solution is the Cauchy Schwarz Inequality, I think it is the nicest way for this inequality.

My solution (maybe it can be similar to some proofs in here ) We have: \[ \left(\frac{a}{a-b}\right)^2+\left(\frac{b}{b-c}\right)^2+\left(\frac{c}{c-a}\right)^2\ge 1\] \[ \Leftrightarrow \frac{(a^2b+b^2c+c^2a-3abc)^2}{(a-b)^2(b-c)^2(c-a)^2}\ge 0\]

Comment $ \boxed{x + y + z = \frac {1}{x} + \frac {1}{y} + \frac {1}{z} = 1\Longrightarrow (x - 1)(y - 1)(z - 1) = 0}\ (xyz\neq 0)$ Here is my solution. Solution 1 By $ xyz = 1$, we have $ \frac {x}{x - 1} + \frac {y}{y - 1} + \frac {z}{z - 1} - \left\{\frac {xy}{(x - 1)(y - 1)} + \frac {yz}{(y - 1)(z - 1)} + \frac {zx}{(z - 1)(x - 1)}\right\}$ $ = \frac {x(y - 1)(z - 1) + y(z - 1)(x - 1) + z(x - 1)(y - 1) - xy(z - 1) - yz(x - 1) - zx(y - 1)}{(x - 1)(y - 1)(z - 1)}$ $ = \frac {x(y - 1)(z - 1 - z) + y(z - 1)(x - 1 - x) + zx(y - 1 - y)}{(x - 1)(y - 1)(z - 1)}$ $ = \frac {x + y + z - (xy + yz + zx)}{(x - 1)(y - 1)(z - 1)}$ $ = \frac {x + y + z - (xy + yz + zx) + xyz - 1}{(x - 1)(y - 1)(z - 1)}$ $ = \frac {(x - 1)(y - 1)(z - 1)}{(x - 1)(y - 1)(z - 1)} = 1\ \because x\neq 1,\ y\neq 1,\ z\neq 1$. $ \therefore \frac {x^2}{(x - 1)^2} + \frac {y^2}{(y - 1)^2} + \frac {z^2}{(z - 1)^2}$ $ = \left(\frac {x}{x - 1} + \frac {y}{y - 1} + \frac {z}{z - 1}\right)^2 - 2\left\{\frac {xy}{(x - 1)(y - 1)} + \frac {yz}{(y - 1)(z - 1)} + \frac {zx}{(z - 1)(x - 1)}\right\}$ $ = \left(\frac {x}{x - 1} + \frac {y}{y - 1} + \frac {z}{z - 1}\right)^2 - 2\left(\frac {x}{x - 1} + \frac {y}{y - 1} + \frac {z}{z - 1} - 1\right)$ $ = \left(\frac {x}{x - 1} + \frac {y}{y - 1} + \frac {z}{z - 1} - 1\right)^2 + 1\geq 1$. The equality holds when $ \boxed{\frac {x}{x - 1} + \frac {y}{y - 1} + \frac {z}{z - 1} = 1 \Longleftrightarrow \frac {1}{x} + \frac {1}{y} + \frac {1}{z} = 3}$. Solution 2 (a) Let $ x + y + z = a,\ xy + yz + zx = b,\ xyz = 1$, $ x,\ y,\ z$ are the roots of the cubic equation : $ t^3 - at^2 + bt - 1 = 0$. If $ t = 1$ is the roots of the equation, then we have $ 1^3 - a\cdot 1^2 + b\cdot 1 - 1 = 0\Longleftrightarrow a - b = 0\ \therefore t\neq 1\Longleftrightarrow a - b\neq 0$. Thus the cubic equation with the roots $ \alpha = \frac {x}{x - 1},\ \beta = \frac {y}{y - 1},\ \gamma = \frac {z}{z - 1}$ is $ \left(\frac {t}{t - 1}\right)^3 - a\left(\frac {t}{t - 1}\right) + b\cdot \frac {t}{t - 1} - 1 = 0$. $ \Longleftrightarrow (a - b)t^3 - (a - 2b + 3)t^2 - (b - 3)t - 1 = 0\ \cdots [*]$. Let $ a - b = p\neq 0,\ b - 3 = q$, we can rerwite the equation as $ pt^3 - (p - q)t^2 - qt - 1 = 0$. By Vieta's formula, we have $ \alpha + \beta + \gamma = \frac {p - q}{p} = 1 - \frac {q}{p},\ \alpha \beta + \beta \gamma + \gamma \alpha = - \frac {q}{p}$. $ \therefore \frac {x^2}{(x - 1)^2} + \frac {y^2}{(y - 1)^2} + \frac {z^2}{(z - 1)^2} = \alpha ^2 + \beta ^2 + \gamma ^2$ $ = (\alpha + \beta + \gamma)^2 - 2(\alpha \beta + \beta \gamma + \gamma \alpha)$ $ = \left(1 - \frac {q}{p}\right)^2 - 2\left( - \frac {q}{p}\right)$ $ = \left(\frac {q}{p}\right)^2 + 1\geq 1$, the equality holds when $ \frac {q}{p} = 0\Longleftrightarrow q = 0\Longleftrightarrow b = 3$, which completes the proof. I really have been enjoying to solve the problem.

I have a proof The ineq can be rewritten in this form $ \sum\frac {1}{(1 - a)^2}\geq\ 1$ there $ x = \frac {1}{a}$ and $ abc = 1$ TO have a nice solution we set $ a + b + c = p,ab + bc + ca = q,abc = r = 1$ we have $ \sum\ (1 - x)^2(1 - y)^2\geq\ {(1 - x)(1 - y)(1 - z)}^2$ if only if $ (p - 3)^2\geq\ 0$ Right for all a,b,c The equality holds if only if $ xyz = 1$ and $ \sum\frac {1}{x} = 3$ the last is solving this system

Solution 3 Here is a natural solution in comparison with the posted solution before. Since $ x,\ y,\ z$ aren't equal to 1, we can set $ x = a + 1,\ y = b + 1,\ z = c + 1\ (abc\neq 0)$. $ \frac {x^2}{(x - 1)^2} + \frac {y^2}{(y - 1)^2} + \frac {z^2}{(z - 1)^2} = \frac {(a + 1)^2}{a^2} + \frac {(b + 1)^2}{b^2} + \frac {(c + 1)^2}{c^2}$ $ = 3 + 2\left(\frac {1}{a} + \frac {1}{b} + \frac {1}{c}\right) + \frac {1}{a^2} + \frac {1}{b^2} + \frac {1}{c^2}$ $ = 3 + \frac {2(ab + bc + ca)}{abc} + \frac {(ab + bc + ca)^2 - 2abc(a + b + c)}{(abc)^2} \cdots [*]$. Let $ a + b + c = p,\ ab + bc + ca = q,\ abc = r\neq 0$, we have $ xyz = 1\Longleftrightarrow p + q + r = 0$. since $ r\neq 0$, we have $ [*] = 3 + \frac {2q}{r} + \frac {q^2 - 2rp}{r^2} = 3 + \frac {2q}{r} + \left(\frac {q}{r}\right)^2 - 2\frac {p}{r}$ $ = 3 + \frac {2q}{r} + \left(\frac {q}{r}\right)^2 + 2\frac {q + r}{r} = \left(\frac {q}{r}\right)^2 + 4\frac {q}{r} + 5$ $ = \left(\frac {q}{r} + 2\right)^2 + 1\geq 1$. The equality holds when $ q = - 2r$ and $ p + q + r = 0\ (r > 0)\Longleftrightarrow p: q: r = 1: ( - 2): 1$. $ Q.E.D.$

y-a-s-s-i-n-e wrote: can_hang2007 wrote: ....... And my solution is the Cauchy Schwarz Inequality, I think it is the nicest way for this inequality. can_hang is the Cauchy schwartz inequality work for negatives ? x , y , z are reals ! I think you dont understand my solution Here is it

Let $a=\frac{x}{1-x}, b=\frac{y}{1-y}, c=\frac{z}{1-z}$. We want to show $a^2+b^2+c^2\ge 1$. We have \[abc=\frac{1}{1-x}\cdot \frac{1}{1-y}\cdot \frac{1}{1-z}=(a+1)(b+1)(c+1)\] Thus, $ab+ac+bc+a+b+c+1=0$. So \begin{align*} (a+b+c+1)^2+1 \\ =a^2+b^2+c^2+2(ab+ac+bc)+2(a+b+c)+2 \\ =a^2+b^2+c^2 \\ \implies a^2+b^2+c^2\ge 1 \end{align*} So part $a$ is done. Equality holds iff $a+b+c+1=0$. We can check that $x=\frac{a}{a+1}$, $y=\frac{b}{b+1}$, $z=\frac{c}{c+1}$. Thus, \begin{align*} xy+yz+xz \\ =\frac{ab}{(a+1)(b+1)}+\frac{bc}{(b+1)(c+1)}+\frac{ac}{(a+1)(c+1)} \\ =\frac{ab(c+1)+bc(a+1)+ac(b+1)}{abc} \\ =\frac{3abc+ab+bc+ac}{abc} \\ =3 \\ \end{align*} Thus we need to show that there exist infinitely many rational $x, y, z$, where $xyz=1$ and $xy+yz+xz=3$. So this is equivalent to $\frac{1}{x}+\frac{1}{y}+xy=3$. Multiplying both sides by $xy$ gives \[y+x+x^2y^2=3xy\implies y^2x^2-(3y-1)x+y=0\] Thus, treating this as a quadratic in $x$ gives a discriminant of $(3y-1)^2-4y^3=(y-1)^2(1-4y)$. Now if we let $y=\frac{1-k^2}{4}$, we get a rational solution for $x$. Now $z=\frac{1}{xy}$ is also rational.

Let $a=\frac{x}{x-1},b=\frac{y}{y-1},c=\frac{z}{z-1}.$ We have $abc=\frac{1}{(x-1)(y-1)(z-1)}.$ Note that $\frac{1}{x-1}=\frac{x}{x-1}-1$ so $abc=(a-1)(b-1)(c-1).$ Thus, $-ab-bc-ca+a+b+c-1=0.$ Now, \[a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=(a+b+c)^2+2-2(a+b+c)=(a+b+c-1)^2+1\ge 1\]as desired. As for the equality case, note that if $a=\frac{x}{x-1}$ then $x=\frac{a}{a-1}$ so putting any rational $a\neq 1$ gives a rational $x.$ Picking any $a+b+c=1$ with rational $a,b,c\neq 1$ gives a rational set of solutions $(x,y,z).$

Let $$a=\frac{x}{x-1},b=\frac{y}{y-1},c=\frac{z}{z-1}.$$From the condition that $xyz=1,$ we have that $$abc=(a+1)(b+1)(c+1)=\frac{1}{1-x}\cdot \frac{1}{1-y}\cdot \frac{1}{1-z}.$$From $abc=(a+1)(b+1)(c+1),$ expanding and arranging, we get $ab+ac+bc+a+b+c+1=0.$ Now, using that equation, we have $$a^2+b^2+c^2+2ab+2ac+2bc+2a+2b+2c+2=(a+b+c+1)^2+1=a^2+b^2+c^2,$$by trivial inequality, $a^2+b^2+c^2 \ge 1.$ So, we finished the first part. For the second part, note that from $a=\frac{x}{x-1},b=\frac{y}{y-1},c=\frac{z}{z-1},$ we can easily get $x=\frac{a}{a+1}, y=\frac{b}{b+1}, z=\frac{c}{c+1}.$ With some computation, we get that $xy+yz+xz=3.$ From here it's just a standard substitutions problem to finish off.

For simplicity set $\tfrac{x^2}{(x-1)^2}=a^2$ etc. Then: $$xyz=1\iff a+b+c=ab+bc+ca+1$$$$\iff 2(a+b+c)=(a+b+c)^2-(a^2+b^2+c^2)+2$$$$\iff a^2+b^2+c^2-1 =(a+b+c-1)^2\geq 0$$And this what we required. Equality when $$a+b+c=1$$

Let $$a=\frac{x}{x-1},b=\frac{y}{y-1},c=\frac{z}{z-1}.$$We have that $$abc=\frac{xyz}{(x-1)(y-1)(z-1)}=\frac{1}{(x-1)(y-1)(z-1)}=(a-1)(b-1)(c-1).$$Let $s_1=a+b+c$ and $s_2=ab+ac+bc.$ Then, we have $s_1=s_2+1$, so $$a^2+b^2+c^2=s_1^2-2s_2=s_2^2+1\geq1.$$ We now wish to show that there are infinitely many rational solutions to $$ab+ac+bc=0, a+b+c=1.$$Solving this, we obtain $$a,b=\frac{-c\pm \sqrt{-3c^2+4c}}{2}.$$Now, we wish to show that there are an infinite number of rationals $c$ such that $-3c^2+4c$ is the square of a rational number. If we let $c=\frac{m}{n}$, then we have $$-3c^2+4c=\frac{-3m^2+4mn}{n^2}.$$Now, it is sufficient for $-3m^2+4mn$ to be a perfect square. Letting $m=1$, it becomes $4n-3$ is a perfect square, and since infinitely many perfect squares are 1 mod 4, we are done.

is this solution correct?

SADAT wrote: is this solution correct? There is a problem in the last line of the proof. x,y,z are real numbers and you cannot use the AM-GM inequality when they are negative numbers,

Generalization 1 $x,y,z,\lambda$ reels ($\lambda \neq 0$) each different from $\sqrt[3]{\lambda ^2}$ hold $xyz=\lambda ^2$. Then prove that $$\sum_{cyc}{\left(\dfrac{x}{x-\sqrt[3]{\lambda ^2}}\right)^2}\geq \lambda ^2+2\sqrt[6]{\lambda^5}\left(1-\sqrt{\lambda}\right)\left(\sum_{cyc}{\dfrac{x}{\sqrt[3]{\lambda^2}}}+1\right)$$

Let $a=\frac{x}{x-1}$. Define $b$ and $c$ similarly. Then \[(a-1)(b-1)(c-1)=abc\implies a+b+c=ab+bc+ca+1.\]Write $a+b+c=s$ and $ab+bc+ca=p$. Then \[a^2+b^2+c^2=s^2-2p=(p^2+2p+1)-2p=p^2+1\ge 1\]with equality when $p=0$ and $s=1$. Then $bc=a^2-a$ and $b+c=1-a$. By the quadratic formula, it suffices for the discriminant $(1-a)(3a+1)$ of the resulting quadratic to be the square of a rational for infinitely many $a$. It would suffice to choose $u$ and $v$ such that \[3u^2+v^2=4\]and then take the value of $a$ such that $1-a=u^2$ and $3a+1=v^2$. Now scale so that $u$ and $v$ are integers, giving the equation \[3u'^2+v'^2=4w^2\]where we want infinitely many distinct $\left(\frac{u'}{w},\frac{v'}{w}\right)$. Notice $u'=4mn$, $v'=6m^2-2n^2$, and $w=3m^2+n^2$ is a valid solution (and can be obtained through difference of squares). Hence take $n=1$ and vary $m$; clearly we get infinite solutions. $\blacksquare$

delegat wrote: (a) Prove that \[\frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} \geq 1\]for all real numbers $x$, $y$, $z$, each different from $1$, and satisfying $xyz=1$. (b) Prove that equality holds above for infinitely many triples of rational numbers $x$, $y$, $z$, each different from $1$, and satisfying $xyz=1$. Author: Walther Janous, Austria We want to prove the 2-variable inequality $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{1}{(xy-1)^2}\geq 1.$$This is true if and only if \begin{align*} & \quad x^4y^4+2x^3y^2+2x^2y^3+x^2+y^2+2xy-6x^3y^3-6x^2y-6xy^2+9x^2y^2 \\ &=(x^2y^2)^2+2x^2y^2(x+y)+(x+y)^2-2(3xy)(x^2y^2+x+y)+(3xy)^2 \\ &=(x^2y^2+x+y-3xy)^2\geq 0\end{align*}this proves (a). Now consider the triple $$(x,y,z)=\left (\frac{2k}{(k+2)^2},-\frac 14k(k+2),-\frac 2{k^2} (k+2)\right),$$where $k$ is any non zero integer. Then $x,y,z\in \mathbb Q \backslash \{1\}$ and $xyz=1$. It is easy to check that equality holds in the above inequality for this triple. This proves (b).

Let $\frac{x}{x - 1} = u$, $\frac{y}{y-1} = v$ and $\frac{z}{z - 1} = w$ so that the condition becomes, \begin{align*} uvw &= (u - 1)(v - 1)(w - 1)\\ \iff uvw &= uvw - uv - vw - wu + u + v + w - 1\\ \iff uv + vw + wu &= u + v + w - 1 \end{align*}and it suffices to show, \begin{align*} u^2 + v^2 + w^2 &\geq 1\\ \iff (u + v + w)^2 - 2(uv + vw + wu) &\geq 1\\ \iff (u + v + w)^2 - 2(u + v + w - 1) &\geq 1\\ \iff (u + v + w - 1)^2 &\geq 0 \end{align*}which is always true. Equality holds exactly when $u + v + w = 1$ so it suffices to show there exist infinitely many solutions to the system, \begin{align*} u + v + w &= 1\\ uv + vw + wu &= 0 \end{align*}Note that the values $u$, $v$ and $w$ are the roots of some polynomial, \begin{align*} P(t) = t^3 - t^2 + b \end{align*}for some $b$. Thus it suffices to show that there exists $b$ such that $t^3 - t^2 + b$ has three distinct roots which follows from looking at the graph of $t^3 - t^2$ and noting it "flips" at two distinct values.

As already pointed out by some people above, P2 of the German MO 2010 was similar. Indeed, the following generalization can also be found here: https://artofproblemsolving.com/community/c6h410315p2298990 Generalization: $\sum_{a,b,c} \left( \frac{ma-nb}{a-b} \right)^2 \ge m^2+n^2 , m,n\in \mathbb{Z}$ is true. The German problem was as terrible as this one. The average score on the problem was only 1/7, similar to P3 (which was in fact IMO 1994/6). I do hope that, if the committee decides to use inequalities in the future, they at least choose some more interesting ones.

doesn't feel like imo... a) Let $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}.$ Then it suffices to show that $$\sum_{cyc} \frac{a^2}{(a-b)^2} \geq 1.$$But expanding everything yields the equivalent inequality $\frac{(a^2b+b^2c+c^2a-3abc)^2}{(a-b)^2(b-c)^2(c-a)^2} \geq 0,$ so we are done. b) Clearly we want $a^2b+b^2c+c^2a=3abc,$ or $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3 \implies xy+yz+zx=3.$ Thus $$xy+\frac{1}{x}+\frac{1}{y}=3 \implies x^2y^2+x+y=3xy \implies y^2x^2+(1-3y)x+y=0.$$This yields a quadratic with discriminant $(1-3y)^2-4y^3=k^2$ for some rational number $p.$ But this factors as $k^2=(y-1)^2(1-4y),$ and thus clearly there are infinitely many rationals $y$ that generate rational $x, z$s, and we are done. QED

Let $a = \frac{x}{x-1}, b = \frac{y}{y-1}, c = \frac{z}{z-1},$ so that $a,b,c$ can take on any real number except for $1.$ Solving for $x,y,z$ gives $x = \frac{a}{a-1}$ and similarly for $y$ and $z.$ We are then given that $$\frac{a}{a-1} \cdot \frac{b}{b-1} \cdot \frac{c}{c-1} = 1 \iff 1 + ab + bc + ca = a + b + c$$upon expansion, and must show that $a^2 + b^2 + c^2 \ge 1,$ with equality holding for infinitely many rational $a,b,c$ (as that would imply that $x,y,z$ are rational, too). Take the given expression and square both sides. We get $(1+ab+bc+ca)^2 = 1 + 2(ab+bc+ca) + (ab+bc+ca)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca.$ This implies that $a^2 + b^2 + c^2 = 1 + (ab+bc+ca)^2 \ge 1,$ finishing part (a). For part (b), one may check that equality in our restatement holds when $$(a,b,c) = \left(\frac{r}{r^2-r+1}, \frac{r^2 - r}{r^2 - r + 1}, \frac{1-r}{r^2- r + 1}\right)$$for any rational $0 < r < 1.$ (We can then substitute for $x,y,z$ according to our definitions of $a,b,c$.) Since there are infinitely many such $r,$ we are done.