Proving that linear combination of exponentials is positive I found the following question in a book without any proof.
Question : Prove that 
$$f(t)=3-5e^{-2t}+6e^{-3t}+2e^{-5t}-3e^{-(3-\sqrt5)t}-3e^{-(3+\sqrt5)t}\gt0$$
for any $t\gt0$.
The book says that this question can be solved in an elementary way. I've tried to prove this, but I'm facing difficulty. Could you show me how to prove this?
 A: The problem in showing that
$$f(t)=3-5e^{-2t}+6e^{-3t}+2e^{-5t}-3e^{-(3-\sqrt5)t}-3e^{-(3+\sqrt5)t}\gt0$$
for all $t > 0$ is that the terms with the slowest decay (except for the constant term) occur with a negative coefficient, that makes it a bit hard to see whether the derivatives are positive everywhere. Simple cure: multiply with a suitable growing exponential, e.g. $e^{5t}$. For
$$g(t) = 3e^{5t} - 5e^{3t} + 6e^{2t} + 2 - 3e^{(2+\sqrt{5})t} - 3e^{(2-\sqrt{5})t},$$
things are easier. We still have $g(0) = 0$ of course, and
$$g'(t) = 15e^{5t} - 15e^{3t} + 12e^{2t} - 3(2+\sqrt{5})e^{(2+\sqrt{5})t} - 3(2-\sqrt{5})e^{(2-\sqrt{5})t},$$
where things still aren't obvious, since the second and third fastest-growing terms have negative coefficients whose sum is larger in absolute value than the coefficient of the fastest-growing term, but further differentiations let the coefficient of the fastest-growing term grow faster than the absolute value of the coefficients of the next fastest-growing terms, so while $g^{(n)}(0) \geqslant 0$, we can simply continue differentiating and get easier to estimate stuff with each differentiation. So, check $g'(0) = 0$, and differentiate once more,
$$g''(t) = 75e^{5t} - 45e^{3t} + 24e^{2t} - 3(9+4\sqrt{5})e^{(2+\sqrt{5})t} - 3(9-4\sqrt{5})e^{(2-\sqrt{5})t}.$$
The coefficient of $e^{5t}$ is still not large enough for a trivial estimate, but we still have $g''(0) = 0$, so let's do it again:
$$g'''(t) = 375e^{5t} - 135e^{3t} + 48e^{2t} - 3(38+17\sqrt{5})e^{(2+\sqrt{5})t} - 3(38-17\sqrt{5})e^{(2-\sqrt{5})t}.$$
Now we're there, $375 > 135 + 3(38+17\sqrt{5})$, so
$$g'''(t) = 3(38+17\sqrt{5})\left(e^{5t} - e^{(2+\sqrt{5})t}\right) + 135(e^{5t} - e^{3t}) + 3(42-17\sqrt{5})e^{5t} + 48e^{2t} + 3(17\sqrt{5}-38)e^{(2-\sqrt{5})t}$$
where each term is easily recognised as non-negative, and even strictly positive for $t > 0$.
So $g''(t)$ is strictly increasing, whence positive, therefore $g'(t)$ is strictly increasing, hence positive, thus $g(t) > 0$ for $t > 0$, and therefore $f(t) > 0$ for $t > 0$.
