Prove $(x_1-x_0)^2+(x_2-x_1)^2+...+(x_n-x_{n-1})^2\leq\frac{1}{n}[(x_1-x_0)^2+(x_2-x_0)^2+...+(x_n-x_0)^2]$

Question

Prove $(x_1-x_0)^2+(x_2-x_1)^2+...+(x_n-x_{n-1})^2\leq\frac{1}{n}[(x_1-x_0)^2+(x_2-x_0)^2+...+(x_n-x_0)^2]$

Could some one help with this?

-----------Edit--------------------------------------

Thank you for the answer of the counter-example. Now I know this inequality does not always hold.

Answer 1

Counterexample: Take $n=2$. Your inequality states

$$(x_2-x_1)^2+(x_3-x_2)^2\leq {1 \over 2}((x_2-x_1)^2+(x_3-x_1)^2).$$

Choose $x_1=x_2\neq x_3.$

Answer 2

Counterexample (one of the infinite out there):
$n = 3$. $x_1 = 5, x_2 = 3, x_3 = 7, x_4 = 3$.
$$(x_2-x_1)^2+(x_3-x_2)^2+(x_4-x_3)^2 \color{red}> \frac{1}{3}((x_2-x_1)^2+(x_3-x_1)^2+(x_4-x_1)^2)$$ since $$(x_2-x_1)^2+(x_3-x_2)^2+(x_4-x_3)^2 = (3-5)^2+(7-3)^2(3-7)^2 = 4+16+16 = 36$$and $$\frac{1}{3}((x_2-x_1)^2+(x_3-x_1)^2+(x_4-x_1)^2) = \frac13((3-5)^2+(7-5)^2+(3-5)^2) = \frac13(4+4+4) = 4$$

Another counterexample where all $x_i$ are distinct (case $n=4$):
$x_1 = 10, x_2 = 8, x_3 = 12, x_4 = 7, x_5 = 13$ I can provide a proof of infinite counterexamples if anyone's interested.

Are there more conditions given?