Proof: It suffices to prove that
$$f(x) = \cos x + a \cos wx - b (\cos wx \cos x + \sin wx \sin x) - (1+a-b)\le 0$$
where $a = (t-1) r^{w-1}, \ b = (1-\frac{1}{t})r^w$. We will use the following facts:
$$
0 < b < 1 < a, \quad \cos \frac{\pi}{w} < 1+2a - 2b - 2ab < 1. \tag{1}
$$
We put the proof at the end.

Clearly, we may restrict to $x\in [0, \pi]$. We split into three cases:

$x \in [\arccos(1+2a-2b-2ab),\, \pi]$: We have
$$f(x)
\le \cos x + \sqrt{(a-b\cos x)^2 + (b\sin x)^2} - (1+a-b)\le 0$$
$$\Longleftarrow (a-b\cos x)^2 + (b\sin x)^2 \le (1+a-b-\cos x)^2$$
$$\Longleftarrow (1-\cos x)(1+2a-2b-2ab-\cos x)\ge 0$$
$$\Longleftarrow \cos x \le 1+2a-2b-2ab$$
$$\Longleftarrow x \in [\arccos(1+2a-2b-2ab), \pi].$$

$x \in [\frac{1}{w}\arccos\frac{a^2 + 2a-2b-2ab}{a^2},\, \arccos(1+2a-2b-2ab))$:
From (1), we have $w\arccos(1+2a-2b-2ab) < \pi$. We have
$$f(x)
\le a\cos wx + \sqrt{(1-b\cos wx)^2 + (b\sin wx)^2} - (1+a-b)\le 0$$
$$\Longleftarrow (1-b\cos wx)^2 + (b\sin wx)^2 \le (1+a-b-a\cos wx)^2$$
$$\Longleftarrow (1-\cos wx)(a^2 + 2a - 2b - 2ab - a^2\cos wx)\ge 0$$
$$\Longleftarrow \cos wx \le \frac{a^2 + 2a-2b-2ab}{a^2}$$
$$ \Longleftarrow
x \in \Big[\frac{1}{w}\arccos\frac{a^2 + 2a-2b-2ab}{a^2},\, \arccos(1+2a-2b-2ab)\Big).
$$

$x\in [0,\, \frac{1}{w}\arccos\frac{a^2 + 2a-2b-2ab}{a^2})$:
From (1), since $a > b$ and $0\le (w-1)x \le wx \le \frac{\pi}{2}$, we have $$f'(x) = -\sin x - aw\sin wx + b(w-1)\sin (w-1)x
\le 0.$$
Since $f(0) = 0$, we have $f(x) \le 0.$ This completes the proof.

$\phantom{2}$

*Proof of (1)*: We only prove $\cos \frac{\pi}{w} < 1 + 2a-2b - 2ab$ because the others are simple.
We will use the following facts:
\begin{align}
1-\frac{2}{w^2}&\ge \cos\frac{\pi}{w}, \qquad\qquad\qquad\qquad\qquad\quad (2)\\
\quad 1 + \frac{1}{tw^2} + \frac{1}{2t^2w^4} &\ge
\Big(1 + \frac{1}{tw^3}\Big)^w. \qquad\qquad\qquad\qquad\ (3)
\end{align}
The proof is not hard and thus omitted.

According to (2), it suffices to prove that $0 \le \frac{1}{w^2} + a-b - ab$ or
$$\frac{1}{w^2} - \frac{t-1}{t} r^{w-1}(r + (t-1)r^w -t) \ge 0.$$
According to (3), noting also that $\frac{t-1}{t} < 1$ and $r + (t-1)r^w -t \ge 1 + (t-1) - t = 0$,
it suffices to prove that
$$\frac{1}{w^2} - \frac{1}{r}\Big(1 + \frac{1}{tw^2} + \frac{1}{2t^2w^4}\Big)
\Big[r + (t-1)\Big(1 + \frac{1}{tw^2} + \frac{1}{2t^2w^4}\Big) -t\Big]\ge 0$$
or
$$\frac{(4 w^6-4 w^5-6 w^4+4 w^3) t^3+(6 w^4-4 w^3-4 w^2) t^2+(4 w^2-2 w-1) t + 1}{4t^3w^5(tw^3+1)}\ge 0$$
which is true since $4 w^6-4 w^5-6 w^4+4 w^3\ge 0, \ 6 w^4-4 w^3-4 w^2 \ge 0, \ 4 w^2-2 w-1 \ge 0.$ This completes the proof of (1).