# Prove/disprove (∫2π0cosf(x)dx)2+(∫2π0√(f′(x))2+sin2f(x)dx)2≥4π2(\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

Let $$f(x)f(x)$$ be a differentiable function on $$[0,2π][0,2\pi]$$ s.t. $$0≤f(x)≤2π0\leq f(x)\leq 2\pi$$ and $$f(0)=f(2π)f(0)=f(2\pi)$$. Prove or disprove that
$$(∫2π0cosf(x)dx)2+(∫2π0√(f′(x))2+sin2f(x)dx)2≥(2π)2 \left(\int_0^{2 \pi} \cos f(x) \,d x\right)^2+\left(\int_0^{2 \pi} \sqrt{(f'(x))^2+\sin^2 f(x)} \, d x\right)^2 \geq(2 \pi)^2$$

It seems that when $$ff$$ is an arbitrary constant, the left side equals $$(2π)2(2\pi)^2$$ and seems to be the minimum. But how can I show that there’s no other $$ff$$ that makes the left side equal (or be less than) $$(2π)2(2\pi)^2$$?

A geometric interpretation of the inequality has been found: Consider a closed curve on a sphere: $$C={(cosx⋅sinf(x),sinx⋅sinf(x),cosf(x))∣x∈[0,2π)}C=\{(\cos x\cdot\sin f(x),\,\sin x\cdot\sin f(x),\,\cos f(x))\mid x\in[0,2\pi)\}$$, we have its perimeter $$L=∫2π0√(f′(x))2+sin2f(x)dx\displaystyle L=\int_0^{2\pi}\sqrt{(f'(x))^2+\sin^2 f(x)}\,dx$$ and its area $$S=2π−∫2π0cosf(x)dx\displaystyle S=2\pi-\int_0^{2\pi}\cos f(x)\,dx$$. From spherical isoperimetric inequality $$L2≥S(4π−S)L^2\ge S(4\pi-S)$$, we have $$(2π−S)2+L2≥(2π)2(2\pi-S)^2+L^2\ge(2\pi)^2$$, and the equality holds iff $$CC$$ is any circle on the sphere. In this way we get the original inequality in the sense of geometry.

Now the question is, how to prove the inequality with only pure analysis methods?

<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 $$A(x)A(x)$$ and $$B(x)B(x)$$, $$(∫βαA(x)dx)2+(∫βαB(x)dx)2≥(∫βα√A(x)2+B(x)2dx)2(\int_{\alpha}^{\beta} A(x) dx)^2+(\int_{\alpha}^{\beta} B(x) dx)^2 \geq (\int_{\alpha}^{\beta} \sqrt{A(x)^2+B(x)^2} dx)^2$$
holds.

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

Let $$A(x)=cosf(x)A(x)=\cos f(x)$$ and $$B(x)=√f′(x)2+sin2f(x)B(x)=\sqrt{{f'(x)}^2+\sin^2 f(x)}$$, and $$α=0,β=2π\alpha=0, \beta=2\pi$$.
Then, by the Prediction, $$(∫2π0cosf(x)dx)2+(∫2π0√(f′(x))2+sin2f(x)dx)2≥(∫2π0√(f′(x))2+1dx)2(\int_{0}^{2\pi} \cos f(x) dx)^2+(\int_{0}^{2\pi} \sqrt{(f'(x))^2+\sin^2 f(x)} dx)^2\geq (\int_{0}^{2\pi} \sqrt{(f'(x))^2+1} dx)^2$$
We know that the right hand side is the form of the length of a curve $$f(x)f(x)$$.
As $$y=f(x)y=f(x)$$ satisfies $$f(0)=f(2π)f(0)=f(2\pi)$$, the shortest length of the curve $$y=f(x)y=f(x)$$ in the interval $$[0,2π][0, 2\pi]$$ would be just a simple line connecting $$(0,f(0))(0, f(0))$$ and $$(2π,f(2π)=f(0))(2\pi, f(2\pi)=f(0))$$, so the length will be just the x-coordinate difference, which is $$2π2\pi$$. Therefore, the RHS will have the minimum of $$(2π)2=4π2(2\pi)^2=4\pi^2$$, 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.
$$(n∑k=0ak)2+(n∑k=0bk)2≥(n∑k=0√(ak)2+(bk)2)2(\sum_{k=0}^{n} a_k)^2+(\sum_{k=0}^{n} b_k)^2 \geq (\sum_{k=0}^{n} \sqrt{(a_k)^2+(b_k)^2})^2$$
I think this equation can be solved by mathematical induction, but I am not sure.