If $x_1=1$ and $x_1,x_2,\ldots,x_{100}$ satisfy the following inequalities:
$$(x_1 - 4x_2 + 3x_3 )\geqslant0\\ (x_2 - 4x_3 + 3x_4 )\geqslant0\\ \vdots\\ (x_{100} - 4x_1 + 3x_2 )\geqslant0$$
Then what is the value of $x_1 + x_2 + \ldots+ x_{100}$?
I couldn't think of anything constructive to do, adding the inequalities does not work and any other steps I did not lead to anything useful.
A hint would do too.
I'm not so sure if I interpreted this problem correctly.
Anyways, summing them all, we see that equality must hold in every inequality.
Now we have $x_i=\frac{x_{i-1}+3x_{i+1}}{4}$, for all $1 \le i \le 100$. (Indices modulo $100$)
Therefore, $$\sum_{i=1}^{100} x^2_i = \sum_{i=1}^{100} (\frac{x_{i-1}+3x_{i+1}}{4})^2 = \sum_{i=1}^{100} (\frac{5}{8}x^2_i + \frac{3}{8} x_{i-1}x_{i+1})$$ This gives $$\sum_{i=1}^{100} x^2_i = \sum_{i=1}^{100} x_{i-1}x_{i+1}$$ This gives $$\sum_{i=1}^{100} (x_{i-1}-x_{i+1})^2 = 0$$ Therefore, $x_{2n+1}=x_1=1$ for all $1 \le n \le 49$.
We also get $x_2=\frac{x_1+3x_3}{4}=\frac{1+3\cdot 1}{4}=1$.
Therefore, $x_{2n}=x_2=1$ for all $1 \le n \le 50$.
The conclusion - $x_i=1$ for all $1 \le i \le 100$, giving the answer of $100$.