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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
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1 Answer

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

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