# 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 : $|\vec{a_1}-\vec{b}|+\cdots+ |\vec{a_n}-\vec{b}| > n$

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Are you talking about...vectors? In the plane? In the space? What have you tried...? – DonAntonio Apr 4 '13 at 2:27
I try $|\overrightarrow{{{a}_{1}}}-\vec{b}|+\cdots +|\overrightarrow{{{a}_{n}}}-\vec{b}|\ge |\overrightarrow{{{a}_{1}}}+\cdots +\overrightarrow{{{a}_{n}}}-n\vec{b}|=|n\vec{b}|>n$，but I use $|\overrightarrow{b}|>1$,not use $|\vec{a_i}|>1$, – geometryscience Apr 4 '13 at 2:42
@DonAntonio, I tried this, that's not enough to get the result. I guess the key is $\sum a_i=0$, but not sure how to use it. And geometryscience's inequality is correct, it is just a generalised version of triangle inequality. – Easy Apr 4 '13 at 3:01
I know it's not enough, @Easy, yet the above inequality that geom. wrote is then enough, without any condition on $\,|a_i|\,$... – DonAntonio Apr 4 '13 at 3:09
The condition given is $|\vec{b}|<1$, not $|\vec{b}|>1$. – pharmine May 13 '13 at 7:24

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 $\vec{y}=\dfrac{\vec{a}}{|\vec{a}|}$.
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 $$\sum_{i=1}^n|\vec{a_i}-\vec{b}|>n.$$