Jtdylktuy

There are ten real numbers $x_0, . . . , x_9$ with $x_0 = 0, x_9 = 9$. What is the smallest possible value of the expression
$$(x_1 − x_0)^2/1 +(x_2 − x_1)^2/2+(x_3 − x_2)^2/3+ · · · +(x_9 − x_8)^2/9$$?
I thought this problem had cauchy schwartz I.E written all over it , I was able to get the $x_i$'s to telescope down to $9^2$ or $ 81$ but then Im getting confused what will be the bi in the sum. will it be the sum of $1+1/2+1/3...+1/9$ or just $1+2+3...+9$ ? please help :(

cauchy schwartz inequality with writing the denominators as a sequence of radicals ]1

As noticed by TheSimpliFire we have

$$(x_1 − x_0)/1 +(x_2 − x_1)/2+(x_3 − x_2)/3+ · · · +(x_9 − x_8)/9=$$$$=(-1)cdot x_0+left(1-frac12right)x_1+left(frac12-frac13right)x_2+ldots+left(frac18-frac19right)x_8+frac19cdot x_9=$$

and by $x_0=0$ and $x_9=9$

$$frac12x_1+frac16x_2+frac1{12}x_3+frac1{20}x_4+frac1{30}x_5+frac1{42}x_6+frac1{56}x_7+frac1{72}x_8+1$$

which can assume any value if $x_1,...,x_8$ are free and if we assume $x_0le x_1le ldots le x_8le x_9$ it is bounded between $1$ and $frac{33}4$.

[![enter image description here][1]][1]

[1]: https://i.stack.imgur.com/fXUVO.jpg by cauchy schwarz the (xi+1 -x i ) telescope leaving 9^2=81 but then we have the denominators which are the sum of integers from 1 to 9 , this gives a total of 9/5