Solution.

(a) Knowing that $|x|+|y| \geqslant|x-y| \Longrightarrow|a+b|+|a+c| \geqslant|a+b-a-c|=|b-c|$. (2p)

(b) Applying, in turn, the relation shown in (a), we obtain

$$\begin{aligned} |x+1|+|x+2014| & \geqslant|x+2014-x-1|=2013 \\ |x+2|+|x+2013| & \geqslant|x+2013-x-2|=2011 \\ \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots & \cdots \cdots+\ldots+1008-x-1007 \mid=1 \end{aligned}$$

Summing the inequalities, we obtain:

$$|x+1|+|x+2|+\ldots+|x+2014| \geqslant 1+3+\ldots+2011+2013=1007^{2}$$

where

$$\begin{aligned} 1+3+\ldots+2011+2013 & =(1+2+\ldots+2014)-(2+4+\ldots+2014) \\ & =\frac{2014 \cdot 2015}{2}-2 \frac{1007 \cdot 1008}{2}=1007^{2} \end{aligned}$$