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?
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 $t > 0$.