Relation between $(c-a)^2 + (d-b)^2$ and $(a-b)^2 + (b-c)^2 + (c-d)^2$

Question

I consider 4 positive numbers (integers) $a< b < c <d.$ I want to know if there exists a relation of the form $$(c-a)^2 + (d-b)^2 \leq (a-b)^2 + (b-c)^2 + (c-d)^2$$

Answer 1

Hint: we get $$(a-b)^2+(b-c)^2+(c-d)^2-(c-a)^2-(d-b)^2=- \left( b-c \right) \left( -b+2\,a+c-2\,d \right) $$

Answer 2

Put $A=(c-a)^2 + (d-b)^2$ and $B=(a-b)^2 + (b-c)^2 + (c-d)^2$.

From $c-a=(c-b)+(b-a)$ and $d-b=((d-c)+(c-b)$ you get

$$A=B-(c-b)(2a-b+c-2d)$$