At any rate, $f$ is non-negative since the cosine takes no value smaller than $-1$, which means that $f$ takes no value smaller than $2 - 1 - 1 = 0$.

Suppose $f(t) = 0$ for some real number $t$, so that
$$0 = (1 + \cos t) + (1 + \cos \sqrt{2} t).$$
But the expression on the right is a sum of two non-negative numbers, hence each of them is zero. So, $1 + \cos t = 1 + \cos \sqrt{2} t = 0$. But this means that there are odd integers $m, n$ such that $t = m\pi$ and $\sqrt{2} t = n\pi$. Clearly, $t \neq 0$, which means that $\sqrt{2}$ is rational, an absurdity. Thus, $f$ is positive.

To prove that $f$ is not periodic, assume the contrary. Then, for some $p > 0$, it is true that $f(t + p) = f(t)$ for all $t$. In particular, $f(p) = f(0) = 4$, equivalently, $2 - \cos p - \cos \sqrt{2} p = (1 - \cos p) + (1 - \cos \sqrt{2} p) = 0$ which means that $\cos p = 1 = \cos \sqrt{2} p$. Thus there are even integers $m, n$ such that $p = m\pi$ and $\sqrt{2} p = n\pi$. But $p \neq 0$. Hence $mn \neq 0$ and yet $m\sqrt{2} = n$, an absurdity.