Prove : $C_{k}(x)+C_{k+1}(x)\geqslant 1$ , $x_{k}\leqslant x\leqslant x_{k+1}$ Suppose $-\infty < x_{1}< x_{2}< \cdots < x_{n}< +\infty (n\geqslant 2)$ . And suppose a algebraic polynomial $C_{k}(x)$ $(k=1,2,\cdots ,n)$  ($degree\leqslant n-1$ ) satisfy :
$$C_{k}(x_{i})=\left\{\begin{matrix} 0, & i\neq k,\\ 1, & i=k, \end{matrix}\right. (i=1,2,\cdots ,n)$$ 

Prove : $C_{k}(x)+C_{k+1}(x)\geqslant 1$ , $x_{k}\leqslant x\leqslant x_{k+1}$ $(1\leqslant k\leqslant n-1)$ .  

Let $C_{k}(x_{i})=a_{k}(x-x_{1})(x-x_{2})\cdots (x-x_{k-1})(x-x_{k+1})\cdots (x-x_{n})$ , $(k=1,2,\cdots ,n)$ .
$a_{k}=\frac{1}{(x-x_{1})(x-x_{2})\cdots (x-x_{k-1})(x-x_{k+1})\cdots (x-x_{n})} $ .
It satisfies the condition . But I can't prove the inequality .
Hope someone helps me . 
Appriciate!
 A: As the answer by Dunham has stated, the solution involves several aspects. First, let
$$P_k(x) = C_k(x) + C_{k+1}(x), \text{ for } 1\leqslant k\leqslant n-1 \tag{1}\label{eq1}$$
As you've indicated, you get
$$C_j(x) = \frac{\prod_{i=1,i\neq j}^{n}(x - x_i)}{\prod_{i=1,i\neq j}^{n}(x_j - x_i)}, \text{ for } 1 \le j \le n \tag{2}\label{eq2}$$
This $n-1$ degree polynomial is the only one of degree $\le n - 1$ which satisfies the required conditions (e.g., as discussed in the answer by Dan in Find N degree polynomial from N+1 points, and with more details in Polynomial interpolation).
As both $C_k(x)$ and $C_{k+1}(x)$ are $n-1$ degree polynomials, their sum of $P_k(x)$ is an up to $n-1$ degree polynomial. Also, $P_k(x_i) = 0$ for all $1 \le i \le n$ except for $i = k$ and $i = k+1$, i.e., for $n-2$ points. Also, note that $P_k(x_k) = P_k(x_{k+1}) = 1$.
Since $P'_k(x)$ is an up to $n-2$ degree polynomial, unless it's the $0$ polynomial, it can have at most $n-2$ roots. However, by the Mean value theorem, $P'_k(x)$ must have a root in each of $(x_i,x_{i+1})$ for all $1 \le i \le n - 1$ except for $i = k-1$ and $i=k+1$. This gives a total of $n-2$ points if $k = 1$ or $k = n - 1$, else $n-3$ points for all other $k$.
Consider that $P_k(x_m) \lt 1$ for some $x_k \lt x_m \lt x_{k+1}$. There are $4$ cases to consider, of which all but the first use the Intermediate value theorem:


*

*For $n = 2$, you get that $P_1(x)$ is a linear function. Since $P_1(x_1) = P_1(x_2) = 1$, this means it must be a constant function of $P_1(x) = 1$, so it's not possible for $P_1(x) \lt 1$.

*For $n \gt 2$ and $k = 1$, since $P_1(x_1) \gt P_1(x_m)$, $P_1(x_2) \gt P_1(x_m)$ and $P_1(x_2) \gt P_1(x_3)$, there are points $x_a$, $x_b$ and $x_c$ where $x_1 \lt x_a \lt x_m$, $x_m \lt x_b \lt x_2$ and $x_2 \lt x_c \lt x_3$ such that $P_1(x_a) = P_1(x_b) = P_1(x_c)$, so by the Mean value theorem, $P'_1(x)$ must at least $2$ roots in $(x_1,x_3)$. Since it has $n-3$ roots in $(x_3,x_n)$, this means $P'_k(x)$ must have at least $2 + (n-3) = n-1$ roots. However, since $P'_k(x)$ can have at most $n-2$ roots, this is not possible.

*For $n \gt 2$ and $1 \lt k \lt n - 2$, you can repeat the argument used in point $2$ above for both sides of $x_m$ to determine there must be at least $3$ roots of $P'_k(x)$ in $(x_{k-1},x_{k+2})$, but as there are $n-4$ other roots, the total of $3 + (n-4) = n - 1$ is once again too large.

*For $n \gt 2$ and $k = n-1$, you can apply the arguments used in point $2$ above for the left side of $x_m$ to, once again, determine that $P'_k(x)$ must have at least $n-1$ roots, which isn't possible.
Since all of the cases for $P_k(x_m) \lt 1$ for some $x_k \lt x_m \lt x_{k+1}$ have been shown to not be possible, this means that $P_k(x) \ge 1$ for all $x_k \le x \le x_{k+1}$ and $1 \le k \le n - 1$ as requested.
A: Hint, rather than a complete answer. Try to prove by contradiction. Sketch some examples with 4-5 points to see where the problem is.
1) $P = C_k + C_{k+1}$ is polynomial of degree $\leq n-1$
2) $P$ has $n-2$ zeros at known locations
3) How many zeros does $P^{\prime}$ have?
