Least $k$ such that $x_1<x_2<x_3<x_4\implies (x_2-x_1)^2 + (x_3-x_2)^2 + (x_4-x_3)^2 + (x_4-x_1)^2 < k\left[ (x_3-x_1)^2 + (x_4-x_2)^2 \right]$

Question

Suppose $x_1,x_2,x_3,x_4\in\mathbb{R}.$ It is not true that

$$x_1<x_2<x_3<x_4\implies (x_2-x_1)^2 + (x_3-x_2)^2 + (x_4-x_3)^2 + (x_4-x_1)^2 < 2\left[ (x_3-x_1)^2 + (x_4-x_2)^2 \right], $$

by consideration of $x_1 = 0, x_2 = 5, x_2 = 6, x_4 = 11,$ we get: $5^2 + 1^2 + 5^2 + 11^2 < 2(6^2 + 6^2),$ which is false.

However, I could not come up with a counterexample for:

$$x_1<x_2<x_3<x_4\implies (x_2-x_1)^2 + (x_3-x_2)^2 + (x_4-x_3)^2 + (x_4-x_1)^2 < \color{blue}{3}\left[ (x_3-x_1)^2 + (x_4-x_2)^2 \right]. $$

Is this true? If yes, is $3$ the best bound for this inequality? If no, is there a best constant bound?

I understand that the inequality is homogeneous and we may suppose WLOG that $x_1,x_2,x_3,x_4\in [0,1],$ or any other interval.

Edit: we do have equality if we consider $x_1 = 0, x_2=x_3=5, x_4=10.$ So maybe we can start from here and say that any deviation of the middle two points $x_2,x_3$ from this scenario doesn't help.

Answer 1

Let $x_2-x_1=a, x_3-x_2=b, x_4-x_3=c$ where $a,b,c>0$.

$$a^2+b^2+c^2+(a+b+c)^2<k((a+b)^2+(b+c)^2)$$ $$\Longleftrightarrow (k-2)a^2+(2k-2)b^2+(k-2)c^2+(2k-2)ab+(2k-2)bc-2ac>0$$ $$\Longleftrightarrow (a-c)^2-(3-k)(a^2+c^2)+(2k-2)(b^2+ab+bc)>0$$

Now it is trivial that this holds when $k=3$.

Suppose $k<3$: let $a=c$ and $b=1/a$ and take $a\rightarrow \infty$. This will make $LHS<0$.

Therefore, $k=3$ is the best bound.

Answer 2

I do not have a full answer, but hopefully this gets you somewhere. Set $$x = x_2 - x_1$$ $$y = x_3 - x_2$$ $$z = x_4 - x_3$$

which leads to the following: $$x^2 + y^2 + z^2 + (x + y + z)^2 < k[(x+y)^2 + (y+z)^2]$$

Define $$f(x,y,z) = \frac{x^2 + y^2 + z^2 + (x + y + z)^2}{(x+y)^2 + (y+z)^2}$$

The task is then to evaluate $$k = \sup \{f(x,y,z) \ |\ x,y,z > 0 \}$$