So we start with a rational $r=\frac{p}{q}$.

1) Say $2\not|p,2\not|q$. Then taking $x=q\pi$, we get:
$$f_r(x)=cos(x)+cos(rx)=cos(q\pi)+cos(p\pi)=-2$$
so $r$ is not the argmax.

2) 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:
$$f_r(x)=-1+cos(\frac{kp}{2q}2\pi)=-1+cos((\frac12+\frac{1}{2q})\pi)=-1-cos(\frac{\pi}{q})$$

For the minimum of $f_r$ to be larger than the minimum of $f_2$, using bounds on $cos$ near $0$, we must have $q\le 2$.

3) Say $q=1$, so also $2|p$. Let $x=(1+\frac{1}{p})\pi$. Then:
$$f_r(x)=-cos(\frac{1}{p}\pi)+cos((p+1)\pi)=-cos(\frac{1}{p}\pi)-1$$

which rules out $p\ge 4$.

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
$$f_r(x)=cos((1/2+\frac{1-2\epsilon}{4k+2})2\pi)+cos((k+\epsilon)\pi)=-1-cos(\frac{1-2\epsilon}{4k+2}2\pi)$$

This rules out the the rest of the possible values. Hence the argmax is at $r=2$.