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matilda
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Is there a geometric or calculus-based reason why the following system of equations should have only one solution?

Let $x_1,x_2,x_3,x_4>0$. Consider the following cyclic system of equations:

$$ 2 + x_2 + x_3 + x_4 + x_2 x_3 x_4 - 2 \left( \frac{x_2}{ \sqrt{x_1 x_2}} + \frac{x_3}{ \sqrt{x_1 x_3}} + \frac{x_4}{ \sqrt{x_1 x_4}} \right)=0,$$

$$ 2 + x_1 + x_3 + x_4 + x_1 x_3 x_4 -2 \left( \frac{x_1}{ \sqrt{x_1 x_2}} + \frac{x_3}{ \sqrt{x_2 x_3}} + \frac{x_4}{ \sqrt{x_2 x_4}} \right)=0,$$

$$ 2 + x_1 + x_2 + x_4 + x_1 x_2 x_4 -2 \left( \frac{x_1}{ \sqrt{x_1 x_3}} + \frac{x_2}{ \sqrt{x_2 x_3}} + \frac{x_4}{ \sqrt{x_3 x_4}} \right)=0,$$

$$ 2 + x_1 + x_2 + x_3 + x_1 x_2 x_3 -2 \left( \frac{x_1}{ \sqrt{x_1 x_4}} + \frac{x_2}{ \sqrt{x_2 x_4}} + \frac{x_3}{ \sqrt{x_3 x_4}} \right)=0.$$

It is long, but easy, to check that this has only one solution, namely $x_1=\dots=x_4=1$.

However, I am not looking for the solution. I am looking for a geometric or calculus-based explanation as to why such a system of equations should have only one solution.

So far, I have the following: every equation has a positive derivative with respect to at least one variable. Hence, if we start at a $0$, we cannot reach another zero by simply moving along any of the coordinate lines. This of course doesn't cover the infinite number of other paths that do not lie along coordinate lines.

matilda
  • 90
  • 10