Prove/disprove $(\int_0^{2 \pi} \!\!\cos f(x) \, d x)^2+(\int_0^{2 \pi}\!\!\! \sqrt{(f'(x))^2+\sin ^2 f(x)} \, dx)^2\ge 4\pi^2$

<This is not an answer!! This is just some of my thoughts, and also I am not good at English. If there are any things to fix, please be my guest.>

Prediction: For two functions $A(x)$ and $B(x)$, $$(\int_{\alpha}^{\beta} A(x) dx)^2+(\int_{\alpha}^{\beta} B(x) dx)^2 \geq (\int_{\alpha}^{\beta} \sqrt{A(x)^2+B(x)^2} dx)^2 $$ holds.

If this is true(which I am not able to prove...), the problem can be solved easily.

Let $A(x)=\cos f(x)$ and $B(x)=\sqrt{{f'(x)}^2+\sin^2 f(x)}$, and $\alpha=0, \beta=2\pi$.
Then, by the Prediction, $$(\int_{0}^{2\pi} \cos f(x) dx)^2+(\int_{0}^{2\pi} \sqrt{(f'(x))^2+\sin^2 f(x)} dx)^2\geq (\int_{0}^{2\pi} \sqrt{(f'(x))^2+1} dx)^2 $$ We know that the right hand side is the form of the length of a curve $f(x)$.
As $y=f(x)$ satisfies $f(0)=f(2\pi)$, the shortest length of the curve $y=f(x)$ in the interval $[0, 2\pi]$ would be just a simple line connecting $(0, f(0))$ and $(2\pi, f(2\pi)=f(0))$, so the length will be just the x-coordinate difference, which is $2\pi$. Therefore, the RHS will have the minimum of $(2\pi)^2=4\pi^2$, and the problem is solved.
However, I couldn't think of a sharp way to prove this Prediction, or even it is true at all. I managed to think of the integrals as sequences, and tried solving this. $$(\sum_{k=0}^{n} a_k)^2+(\sum_{k=0}^{n} b_k)^2 \geq (\sum_{k=0}^{n} \sqrt{(a_k)^2+(b_k)^2})^2$$ I think this equation can be solved by mathematical induction, but I am not sure.

Not an answer (for now), just observations:

We can at least prove that the left-side must be strictly positive. Moreover, this can be done without assuming that $0\leq f(x)\leq 2\pi$ and $f(2\pi)=f(0)$.

Proof: since any sum of squares is at least $0$, the following inequality holds for all differentiable functions $f:[0,2\pi]\to\mathbb{R}$ for which the integrals exist:

$$\left(\int_0^{2\pi}\cos[f(x)]dx\right)^2+\left(\int_0^{2\pi}\sqrt{\left[f'(x)\right]^2+\sin^2\left[f(x)\right]}\text{ }dx\right)^2\geq 0$$

Suppose, for the sake of finding a contradiction, that there is a differentiable $f:[0,2\pi]\to\mathbb{R}$ for which

$$\left(\int_0^{2\pi}\cos[f(x)]dx\right)^2+\left(\int_0^{2\pi}\sqrt{\left[f'(x)\right]^2+\sin^2\left[f(x)\right]}\text{ }dx\right)^2= 0$$

Since $x^2+y^2=0$ if and only if $x=y=0$, it follows that

$$\int_0^{2\pi}\cos[f(x)]dx=0\text{ and }\int_0^{2\pi}\sqrt{\left[f'(x)\right]^2+\sin^2\left[f(x)\right]}\text{ }dx=0$$

From this result together with the observations that $[f'(x)]^2\geq 0$, $\sin^2[f(x)]\geq 0$, and $(\sin\circ f)^2$ is continuous, we can infer that

\begin{align} \int_0^{2\pi}\sqrt{\left[f'(x)\right]^2+\sin^2\left[f(x)\right]}\text{ }dx=0 &\implies \sin^2[f(x)]=0\text{ for every }x\in[0,2\pi]\\ &\implies \sin[f(x)]=0\text{ for every }x\in[0,2\pi]\\ &\implies \text{For every }x\in[0,2\pi]\text{, }f(x)=\pi n\text{ for some }n\in\mathbb{Z} \end{align}

From the continuity of $f$, we deduce that $f=(\pi n)_{[0,2\pi]}$ for some integer $n$ (the notation $c_{[a,b]}$ denotes the constant function mapping $[a,b]$ to $c$). This implies that $\cos\circ f=\pm 1_{[0,2\pi]}$, and consequently

$$\color{red}{0}=\int_0^{2\pi}\cos[f(x)]dx=\int_0^{2\pi}\pm 1dx\color{red}{=\pm 2\pi}$$

a contradiction. Thus, for every differentiable $f:[0,2\pi]\to\mathbb R$,

$$\left(\int_0^{2\pi}\cos[f(x)]dx\right)^2+\left(\int_0^{2\pi}\sqrt{\left[f'(x)\right]^2+\sin^2\left[f(x)\right]}\text{ }dx\right)^2>0$$

Q.E.D

I'm still thinking about how to prove the lower bound of $(2\pi)^2$. Stay tuned :)