Using the inequality $a^2+b^2\geq 2ab$ for $a,b\in\mathbb{R}$ we get $x^2+y^2+z^2+t^2 \geq \frac{1}{4}\cdot(|x-y|^2+|y-z|^2+|z-t|^2+|t-x|^2)$ $\geq [\frac{1}{4}\cdot(|x-y|+|y-z|+|z-t|+|t-x|)]^2=1$ [using the rms mean inequality]

@Carolstar :When does the equality occurs ?

goodar2006 wrote: for nonnegative real numbers $x,y,z$ and $t$ we know that $|x-y|+|y-z|+|z-t|+|t-x|=4$ find the minimum of $x^2+y^2+z^2+t^2$. Using the Cauchy-Schwarz inequality, we have \[\begin{aligned} x^2+y^2+z^2+t^2&=\frac{x^2+y^2}{2}+\frac{y^2+z^2}{2}+\frac{z^2+t^2}{2}+\frac{t^2+x^2}{2} \\ &\ge \frac{(x-y)^2}{2}+\frac{(y-z)^2}{2}+\frac{(z-t)^2}{2}+\frac{(t-x)^2}{2} \\ &\ge \frac{\left( |x-y|+|y-z|+|z-t|+|t-x|\right)^2}{8} =2.\end{aligned}\] On the other hand, it is easy to check equality holds when $x=z=1,$ $y=t=0.$ These two arguments allow us to conclude that \[\min\{x^2+y^2+z^2+t^2\} =2.\, \blacksquare\]

I am extremely sorry. I missed the term 'nonegative'. So I thought $x=z=\frac{1}{2}$ and $y=t=-\frac{1}{2}$ would give the equality.

use this inequality $2x^2+2y^2\geq (x-y)^2$,... and then use Cauchy inequality