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Given the function $$ f(x)=\sum_{i=1}^n c_i e^{-d_i x} $$ where $n\ge 3$ and $d_i\ge 0$, prove or disprove that $f(x)$ has at most three zeros, that is $f(x)=0$ on at most three points.

This looks tedious to answer (no counter-example found). Any hint on how to proceed?

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    One can think of the subcase where $\{ d_1,d_2,\cdots,d_n\}=\{1,2,\cdots,n\}$ as an extension of polynomials, i.e. $f(x)=p(\exp(-x))$ for $p$ some polynomial. It's fairly easy to come up with functions of this form with arbitrarily many zeroes, using this observation. – 2017-01-15

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Note that $$p(x)=(x-1)(x-2)(x-3)=x^3-6x^2+11x-6$$ has three positive zeroes $1,2,3$. Therefore, $$ e^{-3x}-6e^{-2x}+11e^{-x}-6e^{-0x}$$ has zeroes $-\ln1,-\ln 2,-\ln 3$.

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    I made an error in the question, sorry. I meant three zeros! I'm updating the question. – 2017-01-15
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    Well, I guess the same argument applies again. – 2017-01-15