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$|a_1+a_2+ \cdots + a_n|$ comes up a lot when working with polynomials and power series. For the sake of getting an answer, $a_1, ..., a_n$ can be whatever you want; real numbers, complex numbers, positive reals, etc. I've just never seen a lower bound in any of these cases. If we have only two terms, we have the reverse triangle inequality $||a_1|-|a_2|| \le |a_1+a_2|$, but this doesn't generalize to $3$ or more terms.

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    $\begingroup$ let $a_m$ be the minimal element, such that for every $n\in \mathbb{N}: |a_m| \leq |a_n|$. therefore $|a_m| \cdot n \leq |a_1 + a_2 + \cdots + a_n|$ $\endgroup$ – Jneven Mar 20 at 14:34
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    $\begingroup$ @Jneven: Even with absolute values this is not true e.g. $a_1 = 10$ and $a_2 = -10$. $\endgroup$ – MachineLearner Mar 20 at 14:40
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Yes if you have information on arguments (otherwise the answer is trivially no since the sum can cancel); so if for example, assuming $a_k \neq 0, k=1,..n, |a_k| \geq m >0$ and taking arguments in $[-\pi, \pi]$, we have an inequality $|arg(a_k)| \leq \theta < \frac{\pi}{2}$, then $|\Sigma{a_k}| \geq |\Sigma{\Re(a_k)}| \geq mn\cos\theta$.

Similarly, if we have an inequality of the type $\pi - \delta \geq arg(a_k) \geq \delta >0$ as we use the imaginary part, while more generally as long as arguments are contained in an interval of length $a < \pi$, we can rotate all the numbers by the same angle, so keep the absolute value sum same to get into one of the cases above (equivalent by a 90 degree rotation btw) and get a similar inequality

Since arguments that differ by $\pi$ correspond to numbers that cancel when they have the same modulus, we cannot do better than above in general.

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Let us assume we have found a bound $\varepsilon(a_2,\ldots,a_n)$ with $0<\varepsilon(a_2,\ldots,a_n)<|a_1+\cdots+a_n|$. Hence, this bound can depend on all numbers excluding the first number (the choice which number we choose is arbitrary). If we assume that this is the best possible lower bound then we can do the following: we perturb the first number $a_1\to a_1+\delta$. We assume that $0<\delta < \varepsilon$. Then we can write

$$|a_1+\cdots+a_n + \delta|\geq ||a_1+\cdots+a_n|-|\delta||>|\varepsilon-\delta|=\varepsilon-\delta.$$

Hence, it is possible to find an even lower bound if we can at least perturb one of the elements. In the limit, this will lead to $0$ as the final bound.

So in the general case, it is not possible to find a meaningful bound. You will need to add more constraints on the numbers such that you can come up with better bounds. For the complex domain, you could use something like this.

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