The maximal absolute value of one variant in an equation Suppose $n\geq 2$,real numbers $x_i,i=1,2,\ldots,n$ satisfies
$$\sum_{i=1}^n {x_i}^2+\sum_{i=1}^{n-1} x_{i}x_{i+1}=1$$
For each fixed $k,1\leq k\leq n$,show the maximal value of $|x_k|$.
 A: Solution:
Use Cauchy-Schwarz inequality,we have
$$[x^2_{1}+(x_{1}+x_{2})^2+(x_{2}+x_{3})^2+\cdots+(x_{k-1}+x_{k})^2][1+1+\cdots+1]\ge (|x_{1}|+|x_{1}+x_{2}|+\cdots+|x_{k-1}+x_{k}|)^2\ge x^2_{k}$$
because
$$(|x_{1}|+|x_{1}+x_{2}|+\cdots+|x_{k-1}+x_{k}|)^2
\ge (x_{1}-(x_{1}+x_{2})+(x_{2}+x_{3})+\cdots+(-1)^{k-1}(x_{k-1}+x_{k}))^2=x^2_{k}$$
so
$$x^2_{1}+(x_{1}+x_{2})^2+(x_{2}+x_{3})^2+\cdots+(x_{k-1}+x_{k})^2\ge\dfrac{x^2_{k}}{k}\tag{1}$$
the same method,we have
$$(x_{k}+x_{k+1})^2+(x_{k+1}+x_{k+2})^2+\cdots+x^2_{n}\ge\dfrac{x^2_{k}}{n+1-k}\tag{2}$$
$(1)+(2)$ and note
$$x^2_{1}+(x_{1}+x_{2})^2+(x_{2}+x_{3})^2+\cdots+(x_{k-1}+x_{k})^2+(x_{k}+x_{k+1})^2+(x_{k+1}+x_{k+2})^2+\cdots+x^2_{n}=2$$
so we have
$$2\ge \left(\dfrac{1}{k}+\dfrac{1}{n+1-k}\right)x^2_{k}$$
then we have
$$|x_{k}|\le\sqrt{\dfrac{2k(n+1-k)}{n+1}}$$
iff $$x_{1}=-(x_{1}+x_{2})=x_{2}+x_{3}=\cdots=(-1)^{k-1}(x_{k-1}+x_{k})$$
and
$$x_{k}+x_{k+1}=-(x_{k+1}+x_{k+2})=\cdots=(-1)^{n-k}x_{n}$$
or
$$x_{i}=x_{k}(-1)^{i-k}\cdot\dfrac{i}{k},(i=1,2,\cdots,k),\rm{and}.
x_{j}=x_{k}\cdot(-1)^{j-k}\cdot\dfrac{n+1-j}{n-k+1},(j=k+1,k+2,\cdots,n)$$
