So we start with a rational $r=\frac{p}{q}$.
Say $2\not|p,2\not|q$. Then taking $x=q\pi$, we get: $$f_r(x)=cos(x)+cos(rx)=cos(q\pi)+cos(p\pi)=-2$$$$f_r(x)=\cos(x)+\cos(rx)=\cos(q\pi)+\cos(p\pi)=-2$$ so $r$ is not the argmax.
Say $2$ divides exactly one of $p,q$. Then there exists an odd integer $k$ that solves the congruence equation: $$kp=q+1 \pmod{2q}$$ we take $x=k\pi$, and we have: $$f_r(x)=-1+cos(\frac{kp}{2q}2\pi)=-1+cos((\frac12+\frac{1}{2q})\pi)=-1-cos(\frac{\pi}{q})$$$$f_r(x)=-1+\cos(\frac{kp}{2q}2\pi)=-1+\cos((\frac12+\frac{1}{2q})2\pi)=-1-\cos(\frac{\pi}{q})$$
For the minimum of $f_r$ to be larger than the minimum of $f_2$, using bounds on $cos$ near $0$, we must have $q\le 2$.
- Say $q=1$, so also $2|p$. Let $x=(1+\frac{1}{p})\pi$. Then: $$f_r(x)=-cos(\frac{1}{p}\pi)+cos((p+1)\pi)=-cos(\frac{1}{p}\pi)-1$$$$f_r(x)=-\cos(\frac{1}{p}\pi)+\cos((p+1)\pi)=-\cos(\frac{1}{p}\pi)-1$$
which rules out $p\ge 4$.
- Finally, assume $q=2$, so also $2\not|p$. Let $k=\frac{p-1}{2}$, and set $\epsilon=k+1\pmod{2}\in\{0,1\}$. Let $x=\frac{k+\epsilon}{2k+1}2\pi$. Then $$f_r(x)=cos((1/2+\frac{1-2\epsilon}{4k+2})2\pi)+cos((k+\epsilon)\pi)=-1-cos(\frac{1-2\epsilon}{4k+2}2\pi)$$$$f_r(x)=\cos((1/2+\frac{1-2\epsilon}{4k+2})2\pi)+\cos((k+\epsilon)\pi)=-1-\cos(\frac{1-2\epsilon}{4k+2}2\pi)$$
This rules out the the rest of the possible values. Hence the argmax is at $r=2$.
