Yaghi wrote: 1400 real numbers are given. Prove that one can choose three of them like $x,y,z$ such that : $$\left|\frac{(x-y)(y-z)(z-x)}{x^4+y^4+z^4+1}\right| < 0.009$$ Just choose thrice the same number ... .

Quote: Just choose thrice the same number ... . Well actually $x,y,z$ are pairwise distinct. However Yaghi thinks saying "three of them" implies that they are distinct! Even after me and Saeed Agha Haghi had an intense argument with him to convince him that it doesn't, he refused to edit the post; so I decided to correct it in a comment.

the numbers should be different (anyways it wasn't mentioned in the actual exam )

pco wrote: Yaghi wrote: 1400 real numbers are given. Prove that one can choose three of them like $x,y,z$ such that : $$\left|\frac{(x-y)(y-z)(z-x)}{x^4+y^4+z^4+1}\right| < 0.009$$ Just choose thrice the same number ... . you can't choose a number more than once ( not stated on exam paper , but without this assumption , problem would be obvious )

Since dealing with negative numbers is annoying, without loss of generality, assume there are at least $700$ ($513$ is enough) positive numbers among the given $1400$ real numbers \[ 0 < a_1 < \dots < a_{700} = b \]Divide the interval $[0, b]$ into $2^8$ sub-intervals $I_k = \left[\frac{k}{2^8}b, p + \frac{k + 1}{2^8}b \right], 0 \leq k \leq 2^8 - 1$. By the pingeonhole principle, there are $3$ numbers $x > y > z$ belonging to the same interval $I_k$. For these $3$ numbers $x, y, z$ we have \begin{align*} \left| \frac{(x - y)(y - z)(z - x)}{x^4 + y^4 + z^4 + 1} \right| &\leq (x - y)(y - z)(x - z) \leq \frac{1}{4}(x - z)^3\\ &\leq \frac{1}{4} \left( \frac{b}{2^8} \right)^3 = \frac{b^3}{2^{26}} \end{align*}If $b \leq 64$ then the desired quantity does not exceed $\frac{64^3}{2^{26}} = 0.0390625$ and we are done. Otherwise $b > 64$ and by choosing $x = a_{700} = b$ and $y > z$ arbitrary we have \begin{align*} \left| \frac{(x - y)(y - z)(z - x)}{x^4 + y^4 + z^4 + 1} \right| &= \frac{(b - y)(y - z)(b - z)}{b^4 + y^4 + z^4 + 1} \\ &\leq \frac{ \frac14 b^3}{b^4} < \frac{1}{4b}\\ &< \frac{1}{4 \times 64} = 0.00390625 \end{align*} PS: I think we can get much tighter bounds, anyone has any ideas?

A logical way... If we have two equal numbers, the statement will be obvious. On the other hand, we can assume that at least $700$ of them are positive. Suppose $x> y > z \geq 0$ Now the equation will become: $$\frac{(x-y)(y-z)(x-z)}{x^4+y^4+z^4+1} < 0.009$$$$\iff 9(x^4+y^4+z^4+1)>1000(x-y)(y-z)(x-z)=1000(x^2y-x^2z-xy^2+xz^2-yz^2+y^2z)$$$$\iff 9(x^4+y^4+z^4+1)+1000(x^2z+xy^2+yz^2)>1000(x^2y+xz^2+y^2z)$$We know that: $$xy^2>xz^2 , x^2z>y^2z$$So it is enough to prove: $$9(x^4+y^4+z^4+1)+1000yz^2>1000x^2y$$$$\iff x^2(9x^2-1000y)+z^2(9z^2+1000y)+9y^4+9>0$$If $ \exists x$ such $x\geq \frac{1000}{9}\to 9x^2>1000y$ and the inequality will be obvious. Now consider all $700$ numbers are less than $\frac{1000}{9}$ : $$a_1<a_2<a_3<.....<a_{700}<\frac{1000}{9}$$Let $c=a_{2k+1}-a_{2k-1}$ be the $min$ between following $349$ expressions: $$a_3-a_1,a_5-a_3,\dots,a_{699}-a_{697}$$Consider $x>y>z$ as $a_{2k+1}>a_{2k}>a_{2k-1}$ we'll have: $$x-z\leq c, (y-z)(x-y)\leq \frac{c^2}{4} AM-GM$$$$\to \frac{(x-y)(y-z)(x-z)}{x^4+y^4+z^4+1} < \frac{c^3}{4}$$Now thats enough to have $\frac{c^3}{4}\leq \frac{9}{1000} \iff c\leq \frac{\sqrt[3]36}{10}$ For the sake of contradiction, let us assume: $c>\frac{\sqrt[3]36}{10}$: $$\to a_{699}=a_1+(a_3-a_1)+(a_5-a_3)+.....+(a_{699}-a_{697})\geq a_1 + 349c $$$$\geq a_1 + 349\frac{\sqrt[3]36}{10} > 115.23$$We arrive at a contradiction: $\frac{1000}{9}~111.11>a_{699}>115.23$

@mhr2004 Can you obtain a better upper bound?

Is there a generalisation with n- reals numbers ?

@Moubinool It depends on what kind of generalizations you want. For example, using my method we can show that among $2^{n + 1} + 1$ positive real numbers we can choose $3$ numbers $x, y, z$ satisfying \[ \left| \frac{(x - y)(y - z)(z - x)}{x^4 + y^4 + z^4 + 1} \right| < \frac{1}{2^{\frac34 n + 2}} \] This bound is, in my opinion, extremely weak but unfortunately I don't know how to do better.

After discussing with @mhr2004, I managed to lower the upper bound to $\frac{1}{2000} = 0.0005$. The most crucial idea of this proof is his. I'm reusing the notations from my post #6. Also denote $f(x, y, z)$ the given expression for brevity. For starter, if $b \geq 500$ then by the estimate I established at the end of post #6, \[ f(b, y, z) < \frac{1}{4b} \leq \frac{1}{4 \times 500} = \frac{1}{2000} \]so from now on, consider $b < 500$. If $b \geq 160$, there are two sub-cases to consider $\star$ There are two other numbers $y > z$ in the interval $\left[ \frac{8}{25}b, b \right]$. In this case \[ f(b, y, z) < \frac{\frac14 \cdot \left(\frac{17b}{25} \right)^3}{b^4 + \left(\frac{8}{25}b\right)^4 + \left(\frac{8}{25}b\right)^4} = \frac{122825}{1595268b} < \frac{122825}{1595268 \times 160} \approx 0.0004812 \] $\star$ At least $699$ numbers belongs to the interval $\left[0, \frac{8}{25}b \right)$ whose length is $\frac{8}{25}b < 160$ Thus in the case $b < 500$, we can always suppose that at least $698$ numbers $a_1, \dots, a_{698}$ belong to $[0, 160)$. Now consider 160 intervals $[0, 1), [1, 2), \dots, [159, 160)$. If there are $5$ numbers in some interval $[k, k + 1)$, $k \geq 3$ then at least $3$ numbers lie in either $\left[k, k + \frac12 \right)$ or $\left[k + \frac{1}{2}, k + 1 \right)$. In either cases we can bound \[ f(x, y, z) \leq \frac{ \frac14 \cdot \left( \frac12 \right)^3 }{3k^4 + 1} \leq \frac{ \frac14 \cdot \left( \frac12 \right)^3 }{3 \cdot 3^4 + 1} \approx 0.000128 \] Otherwise in each interval $[k, k + 1)$, $3 \leq k \leq 159$ there are at most $4$ numbers. So there are at least $698 - 157 \times 4 = 70$ numbers distributed among the three intervals $[0, 1), [1, 2), [2, 3)$. Thus there is an interval containing at least $24$ numbers, from which we can choose three $x > y > z$ belonging to a sub-interval of length $\frac{1}{8}$. In this case \[ f(x, y, z) < \frac{1}{4} \cdot \left( \frac{1}{8} \right)^3 \approx 0.000488 \]so we are done!

1)There are $x, y, z$ if the following condition is met $\frac{9}{1000} \geq |(x-y)(y-z)(z-x)|$ $\frac{9}{1000} \geq \frac{|(x-y)(y-z)(z-x)|}{1} \geq \frac{|(x-y)(y-z)(z-x)|}{x^4+y^4+z^4+1}$ 2)So for every $x, y, z$ we have that: $|(x-y)(y-z)(z-x)| \geq \frac{9}{1000}$ $a_{1400} \geq a_{1399} \geq a_{1398} \geq.... \geq a_{1}$ $a_{1400}-a_{1399},a_{1399}-a_{1398},a_{1398} -a_{1397},........,a_{2} -a_{1}$ $a_{1400}-a_{1399},a_{1399}-a_{1398},a_{1400}-a_{1398}$,It is not possible for all three expressions to be smaller than $\sqrt[3] {\frac{9}{1000}}$ $a_{1400}-a_{1398} \geq \sqrt[3] {\frac{9}{1000}}$,a$_{1398}-a_{1396} \geq \sqrt[3] {\frac{9}{1000}}$,$a_{1396}-a_{1394} \geq \sqrt[3] {\frac{9}{1000}}$,.....,$a_{4}-a_{2} \geq \sqrt[3] {\frac{9}{1000}}$ Collect all the above phrases $a_{1400}-a_{2} \geq 699\sqrt[3] {\frac{9}{1000}} \geq 140$ So one of two numbers $|a_{1400}|$ , $|a_{2}|$ bigger than $70$ In a similar way, it can be concluded that $|a_1|,|a_2|,|a_3|\geq 69$ or $|a_{1400}|,|a_{1399}|,|a_{1398}|\geq 69$ 1) $a_{1400} \geq a_{1399} \geq a_{1398} \geq 69$ $x \geq y \geq z$ $\frac {(x-z)^3}{x^4+y^4+z^4+1} \geq \frac{|(x-y)(y-z)(z-x)|}{x^4+y^4+z^4+1}$ Then $x=a_{1400}, y=a_{1399},z=a_{1398}$ if $2a_{1398} > a_{1400}$,....... End problem otherwise $a_{1400}>138$: $\frac{9}{1000}>\frac{1}{138}>\frac{1}{x}>\frac {(x-z)^3}{x^4+y^4+z^4+1}$ 2)$a_{1400}, a_{1399},a_{1398}$ is negative $|a_{1400}|,|a_{1399}|,|a_{1398}|> 69$ $a_{1400}, a_{1399}, a_{1398}<-69$ $a_{1400}-a_{2}>140$,$-a_{2}>209,a_{2}<-209$ Then Same way $a_{1},a_{2},a_{3}<-208$ $x=a_{1}, y=a_{2}, z=a_{3}$ kill problem..... And the desired result is achieved. 3)$a_{1400}, a_{1399},a_{1398}$ is negative $|a_{3}|,|a_{2}|,|a_{1}|> 69$........ $a_{1}, a_{2}, a_{3}<-69$ because $a_{1400}-a_{2}>140$.....$a_{1400}>140+a_{2}>71$ then There is a contradiction. 4)$a_{1400}$ is postive, $a_{1398}$ is negative Very similar to previous cases

$x_1<x_2<...<x_{1400} \longrightarrow_{there exist x,y,z} \left|\frac{(x-y)(x-z)(y-z)}{x^4+y^4+z^4+1} \right|<0.009$ We claim that if three of the numbers, such as $x_i,x_{i+1},x_{i+2}$ belong to the interval ($t,t+0.3$)the question will be concluded.because: $\frac{(x_{i+2}-x_{i+1})(x_{i+1}-x_i)(x_{i+2}-x_i)}{x_i^4+x_{i+1}^4+x_{i+2}^4+1}<(x_{i+2}-x_{i+1})(x_{i+1}-x_i)(x_{i+2}-x_i))\le (\frac{(x_{i+2}-x_{i+1}+x_{i+1}-x_i}{2})^2(x_{i+2}-x_i)=\frac{1}4(x_{i+2}-x_i)^3<\frac{(0.03)^3}{4}<\frac{9}{1000}$ hence Therefore, the distance between numbers is one among the at least $0.3$. According to the Pigeonhole principle, at least $700$ numbers are negative or positive.WLOG we suppose that: $$0\leq x_{700}<x_{701}<...<x_{1400}$$Hence $0.3<x_{702},0.6<x_{704},...,699\cdot 0.3<x_{1398}$ $\Rightarrow x_{1398},x_{1399},x_{1400}\in (200,\infty)$ We prove that for these numbers is true.because $\frac{(x_{1400}-x_{1399})(x_{1399}-x_{1398})(x_{1400}-x_{1398})}{x_{1400}^4+x_{1399}^4+x_{1398}^4+1}<\frac{1}{4}\frac{(x_{1400}-x_{1398})^3}{x_{1400}^4}<\frac{1}{4}\frac{x_{1400}^3}{x_{1400}^4}=\frac{1}{4x_{1400}}<\frac1{800}<\frac1{900}$

Hopeooooo wrote: $x_1<x_2<...<x_{1400} \longrightarrow_{there exist x,y,z} \left|\frac{(x-y)(x-z)(y-z)}{x^4+y^4+z^4+1} \right|<0.009$ We claim that if three of the numbers, such as $x_i,x_{i+1},x_{i+2}$ belong to the interval ($t,t+0.3$)the question will be concluded.because: $\frac{(x_{i+2}-x_{i+1})(x_{i+1}-x_i)(x_{i+2}-x_i)}{x_i^4+x_{i+1}^4+x_{i+2}^4+1}<(x_{i+2}-x_{i+1})(x_{i+1}-x_i)(x_{i+2}-x_i))\le (\frac{(x_{i+2}-x_{i+1}+x_{i+1}-x_i}{2})^2(x_{i+2}-x_i)=\frac{1}4(x_{i+2}-x_i)^3<\frac{(0.03)^3}{4}<\frac{9}{1000}$ hence Therefore, the distance between numbers is one among the at least $0.3$. According to the Pigeonhole principle, at least $700$ numbers are negative or positive.WLOG we suppose that: $$0\leq x_{700}<x_{701}<...<x_{1400}$$Hence $0.3<x_{702},0.6<x_{704},...,699\cdot 0.3<x_{1398}$ $\Rightarrow x_{1398},x_{1399},x_{1400}\in (200,\infty)$ We prove that for these numbers is true.because $\frac{(x_{1400}-x_{1399})(x_{1399}-x_{1398})(x_{1400}-x_{1398})}{x_{1400}^4+x_{1399}^4+x_{1398}^4+1}<\frac{1}{4}\frac{(x_{1400}-x_{1398})^3}{x_{1400}^4}<\frac{1}{4}\frac{x_{1400}^3}{x_{1400}^4}=\frac{1}{4x_{1400}}<\frac1{800}<\frac1{900}$ this is indeed a magnificent discovery$\frac{1}{800} \le \frac{1}{900}$

Indeed it is