Question

Given `{r,s,u,v} in RR^4` Prove that `min {r-s^2,s-u^2,u-v^2,v-r^2} le 1/4`?

Answer 1

See explanation...

Explanation:

First note that if `r = s = u = v = t` then we are seeking to maximise:

`t-t^2 = 1/4-(1/4-t+t^2) = 1/4-(t-1/2)^2`

so the maximum `1/4` occurs when `t = 1/2`

Let:

`{ (a = r-1/2), (b = s-1/2), (c = u-1/2), (d = v-1/2) :}`

Then:

`{ (r-s^2-1/4 = a+1/2 - (b+1/2)^2 - 1/4 = a-b-b^2), (s - u^2-1/4 = b+1/2 - (c+1/2)^2 - 1/4 = b-c-c^2), (u - v^2-1/4 = c+1/2 - (d+1/2)^2 - 1/4 = c-d-d^2), (v - r^2-1/4 = d+1/2 - (a+1/2)^2 - 1/4 = d-a-a^2) :}`

Note that `a^2 >= 0`, `b^2 >= 0`, `c^2 >= 0` and `d^2 >= 0`, so if all of the above expressions are non-negative, then:

`a >= b >= c >= d >= a`

Hence `a=b=c=d` and we are back to `r = s = u = v`