# If $(x_i^2-x_{i+2}x_{i-1})(x_{i+1}^2-x_{i+2}x_{i-1})\leq 0$ for all $i=1,2,3,4,5$ (indices considered mod $5$), then $x_1=x_2=x_3=x_4=x_5$.

Question. Let $$x_1,x_2,x_3,x_4,x_5>0$$. If $$(x_1^2-x_3x_5)(x_2^2-x_3x_5)≤0,$$ $$(x_2^2-x_4x_1)(x_3^2-x_4x_1)≤0,$$ $$(x_3^2-x_5x_2)(x_4^2-x_5x_2)≤0,$$ $$(x_4^2-x_1x_3)(x_5^2-x_1x_3)≤0,$$ and $$(x_5^2-x_2x_4)(x_1^2-x_2x_4)≤0,$$ then prove that the only solution to these inequalities is $$x_1=x_2=x_3=x_4=x_5=a$$ where $$a$$ is some real number.

What I tried
Initially I assumed that $$0

Using the assumption that $$0 and the given inequality $$(x_1^2-x_3x_5)(x_2^2-x_3x_5)≤0$$ We get $$x_1^2≤x_3x_5$$ and $$x_2^2≥x_3x_5$$ Upon using all inequalities in a similar way, it becomes clear that there is a contradiction unless $$x_1=x_2=x_3=x_4=x_5=a$$ is true

So initially I thought I had proved it.

Since all the inequalities seemed to be symmetric for $$x_1,x_2,x_3,x_4$$ and $$x_5$$, it did not matter that I had assumed $$0 to prove it.

However on closer inspection, I realised that the equations aren't truly symmetric. For example we have $$(x_1^2-x_3x_5)$$ and $$(x_1^2-x_2x_4)$$ but not $$(x_1^2-x_3x_2)$$ and other such terms used in the inequalities. I wasn't able to figure out if this means that I cannot assume $$0.

So I would like to know

1)Is my assumption is valid?

2)If it isnt valid (or even if it is), what is an alternate solution to this problem?

Thank you so much in advance

Regards

Your solution isn't far from mine... so I show you. The idea is based on factorizations.

We can write: \begin{align} 2(x_1^2 - x_3x_5)(x_2^2 - x_3x_5) & \leq 0 \iff x_1^2x_2^2+x_1^2x_2^2-2x_1^2x_3x_5-2x_2^2x_3x_5+x_3^2x_5^2+x_3^2x_5^2 \le 0, \\ 2(x_2^2 - x_4x_1)(x_3^2 - x_4x_1) & \leq 0\iff x_2^2x_3^2+x_2^2x_3^2-2x_2^2x_4x_1-2x_3^2x_4x_1+x_4^2x_1^2+x_4^2x_1^2 \le 0,\\ 2(x_3^2 - x_5x_2)(x_4^2 - x_5x_2) & \leq 0\iff x_3^2x_4^2+x_3^2x_4^2-2x_3^2x_5x_2-2x_4^2x_5x_2+x_5^2x_2^2+x_5^2x_2^2 \le 0,\\ 2(x_4^2 - x_1x_3)(x_5^2 - x_1x_3) & \leq 0\iff x_4^2x_5^2+x_4^2x_5^2-2x_4^2x_1x_3-2x_5^2x_1x_3+x_1^2x_3^2+x_1^2x_3^2 \le 0,\\ 2(x_5^2 - x_2x_4)(x_1^2 - x_2x_4) & \leq 0\iff x_1^2x_5^2+x_1^2x_5^2-2x_5^2x_2x_4-2x_1^2x_2x_4+x_2^2x_4^2+x_2^2x_4^2 \le 0. \end{align}

So, \begin{align} \color{\orange}{x_1^2x_2^2}+x_1^2x_2^2-\color{\lightblue}{2x_1^2x_3x_5}-\color{\red}{2x_2^2x_3x_5}+\color{\pink}{x_3^2x_5^2}+\color{\purple}{x_3^2x_5^2} \le 0 \\ \color{\red}{x_2^2x_3^2}+\color{\pink}{x_2^2x_3^2}-\color{\orange}{2x_2^2x_4x_1}-\color{\grey}{2x_3^2x_4x_1}+\color{\maroon}{x_4^2x_1^2}+x_4^2x_1^2 \le 0\\ \color{\grey}{x_3^2x_4^2}+\color{\maroon}{x_3^2x_4^2}-\color{\pink}{2x_3^2x_5x_2}-\color{\green}{2x_4^2x_5x_2}+\color{\magenta}{x_5^2x_2^2}+\color{\red}{x_5^2x_2^2} \le 0\\ \color{\green}{x_4^2x_5^2}+\color{\magenta}{x_4^2x_5^2}-\color{maroon}{2x_4^2x_1x_3}-\color{\purple}{2x_5^2x_1x_3}+\color{\lightblue}{x_1^2x_3^2}+\color{\grey}{x_1^2x_3^2} \le 0\\ \color{\purple}{x_1^2x_5^2}+\color{\lightblue}{x_1^2x_5^2}-\color{\magenta}{2x_5^2x_2x_4}-2x_1^2x_2x_4+\color{\orange}{x_2^2x_4^2}+\color{\green}{x_2^2x_4^2} \le 0 \end{align} where each color corresponds to a factorization.

When we add the colored inequalities, we get: \begin{align}&(x_1x_2-x_1x_4)^2+(x_2x_3-x_2x_5)^2+(x_3x_4-x_3x_1)^2+(x_4x_5-x_4x_2)^2\\&\ \ \ \ \ \ \ \ \ \ +(x_5x_1-x_5x_3)^2+(x_1x_3-x_1x_5)^2+(x_2x_4-x_2x_1)^2+(x_3x_5-x_3x_2)^2\\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +(x_4x_1-x_4x_3)^2+(x_5x_2-x_5x_4)^2 \le 0.\end{align} So all there squares equal $$0$$. So $$x_1=x_2=x_3=x_4=x_5.$$