Given a function $$f(t) = k_{1} \sin(t+\alpha) + k_{3} \sin(3t+\beta) + k_{5} \sin(5t+\gamma)$$ where $k_{1}, k_{3}$ and $k_{5}$ are all positive parameters, and the three phase angles, $ 0<\alpha<\pi/2, 0<\beta,\gamma<2\pi$. The problem is: if $f(t)$ has only two extreme points in a period $[0,2\pi]$, say $t_1$, $t_2 \in [0,2\pi]$, due to anti-symmetry of $f(t)$, without loss of generality，$t_2$ = $t_1 + \pi$， then does the following inequality always hold for $k_{1}, k_{3}$ and $k_{5}$:

$$k_{1}> k_{3} > k_{5}$$

My first impression was to use time derivative of $f(t)$ and root of $f(t)$, i.e.,

$$f'(t_1) = k_{1}\cos(t_1+\alpha) + 3k_{3} \cos(3t_1+\beta) + 5k_{5} \cos(5t_1+\gamma)=0$$ $$f(t_1+\pi/2) = k_{1}\cos(t_1+\alpha) -k_{3}\cos(3t_1+\beta) + k_{5} \cos(5t_1+\gamma)=0$$ but I have no idea how to go further. Maybe I need some additional constraints for the phase angle parameter. I need some help here, please.

P.S. I've seen that the pattern of such relation $$k_{1}> k_{3} > k_{5}$$ is true in nearly all cases that I've encountered, but I have never seen any theoretical work as to why this is the case.