Prove that if $\sum_{i=1}^n\lambda_i|x-a_i|=0$ then $\lambda_1=\lambda_2=\cdots \lambda_n=0$ Let $ n\in \Bbb N $ and $ (a_1,a_2,\cdots,a_n)\in \Bbb R^n $ such that
$$a_1<a_2<\cdots <a_n$$
Prove that if there exist $(\lambda_1,\lambda_2,\cdots,\lambda_n)\in \Bbb R^n $ satifying
$$(\forall x\in\Bbb R)\; \sum_{i=1}^n\lambda_i|x-a_i|=0$$
then
$$\lambda_1=\lambda_2=...=\lambda_n=0$$
I tried induction in vain. Any idea will be appreciated.

Denote $P(x)$ as the function $\sum_{i=1}^n \lambda_i |x-a_i|$.
For each $i\in [n]$, there exists some $\epsilon_i>0$ such that $a_i-\epsilon_i>a_{i+1}$ and $a_i+\epsilon_i=a_{i+1}$ (where we can treat $a_0$ and $a_{n+1}$ as very small and large numbers). For example something like $\epsilon_i=\text{min}\left(\frac{a_{i+1}-a_i}{2},\frac{a_i-a_{i-1}}{2}\right)$. We can deduce that
$$P(a_i+\epsilon_i)-P(a_i-\epsilon_i)=2\lambda_i\epsilon_i=0$$
So it follows that $\forall i\in [n], \lambda_i=0$
