Prove $ |\vec{a_1}-\vec{b}|+ \cdots +|\vec{a_n}-\vec{b}| > n $ There are $ \vec{a_1},\vec{a_2},\vec{a_3}, \ldots ,\vec{a_n},\vec{b}\; $ such that $ |\vec{a_i}|>1 $, $ |\vec{b}|<1 $, $ \vec{a_1}+\cdots+\vec{a_n}=0 $
.Prove  : $ |\vec{a_1}-\vec{b}|+\cdots+ |\vec{a_n}-\vec{b}| > n $
 A: Lemma. Let $\vec{a}$ be a nonzero vector and $\vec{b}$ be another vector. Then $|\vec{a}-\vec{b}|\ge |\vec{a}|-\vec{b}\cdot\dfrac{\vec{a}}{|\vec{a}|}$. 
Proof of Lemma. Use $|\vec{x}||\vec{y}|\ge \vec{x}\cdot\vec{y}$ for  $\vec{x}=\vec{a}-\vec{b}$ and $\vec{y}=\dfrac{\vec{a}}{|\vec{a}|}$.
Proof. Using the above lemma, we get
\begin{align*}
&\sum_{i=1}^n|\vec{a_i}-\vec{b}|\\
&\ge \sum_{i=1}^n \left(|\vec{a_i}|-\vec{b}\cdot\frac{\vec{a_i}}{|\vec{a_i}|}\right)\\
&=\sum_{i=1}^n \left(1-\vec{b}\cdot\vec{a_i}\right)+\sum_{i=1}^n\left(|\vec{a_i}|-1+\vec{b}\cdot\frac{\vec{a_i}}{|\vec{a_i}|}\cdot (|\vec{a_i}|-1)\right)\\
&=n-\vec{b}\cdot\sum_{i=1}^n \vec{a_i}+\sum_{i=1}^n\frac{|\vec{a_i}|-1}{|\vec{a_i}|}(|\vec{a_i}|-\vec{b}\cdot\vec{a_i}).
\end{align*}
Here, the second term $\vec{b}\cdot\sum_{i=1}^n \vec{a_i}$ is zero. The third term is positive because $|a_i|>1$ and
$$|\vec{a_i}|>|\vec{b}||\vec{a_i}|\ge \vec{b}\cdot\vec{a_i}.$$
The first inequality follows from $|b|<1$. Hence
$$\sum_{i=1}^n|\vec{a_i}-\vec{b}|>n.$$
