I am trying to prove the following inequality but have no idea at all. Does anyone have any idea how to tackle it or know some related sources? $$\frac{\int_0^{2\pi}(\frac{1}{2}\frac{f^2}{g}\frac{dg}{ds}+3f\frac{df}{ds})^2g+v(-\frac{1}{v}f^3g+\frac{d^2f}{ds^2})^2+kvf^6g\ ds}{\int_0^{2\pi}f^4g+v(\frac{df}{ds})^2\ ds}\geq \min(1,kv)$$ where $f,g$ are smooth positive periodic functions, $\int_0^{2\pi}f\ ds=2\pi$, and $k,v$ are positive constants.
It arises from a geometric problem. It is controlled by $1$ if let $g$ tends to the zero function, while controlled by $kv$ if $f=1$. I have tried some examples in python and it does seem that $min(1,kv)$ should be a valid bound. Indeed any bound independent of $f,g$ would be useful for my purpose, $min(1,kv)$ is my conjectured optimal bound. Any help would be appreciated.
Some updates:
A counterexample is already given by @abacaba below, but I still wonder if some other bound might exist (could depend on k,v, but not on f,g).
If we change the term $(\frac{1}{2}\frac{f^2}{g}\frac{dg}{ds}+3f\frac{df}{ds})^2g$ in the numerator to $(\frac{1}{2}\frac{f^2}{g}\frac{dg}{ds}+\alpha f\frac{df}{ds})^2g$ for any $\alpha\in (-\infty,\frac{3}{2}),$ then a valid bound would be $$\sup_{\epsilon\in(0,\frac{1}{2\pi}),\rho\in\mathbb{R}^+}\min(\frac{(1-\sqrt{2\pi\epsilon})^6}{(1+\sqrt{2\pi\epsilon})^4}kv,\frac{\delta}{\rho+1},\sqrt{\frac{2\delta\rho\epsilon kv}{\pi}}),$$ where $\delta=min(\frac{3}{2}-\alpha,1).$
On the other hand, for $f$ very close to a constant function, in the sense that $\sup_{x\in[0,2\pi]} |\frac{df}{ds}|^2(x)<\mu$, where $\mu$ is some small constant only depends on $v$, a valid bound would be $$\min(\frac{(1-2\pi\sqrt{\frac{\mu}{v}})^6}{(1+2\pi\sqrt{\frac{\mu}{v}})^4}kv,1)-10\frac{(1+2\pi\sqrt{\frac{\mu}{v}})^2}{(1-2\pi\sqrt{\frac{\mu}{v}})^4}\frac{\mu}{v}.$$ Notice that for $\mu$ small enough, the above bound is always positive.
I think this inequality might not be true.
First take $g = f^{-6}$ and $kv = 1$. Then the first term in the numerator vanishes, so we are left with $$\int_0^{2\pi} v\left(-\frac{1}{v} f^{-3} + \frac{d^2 f}{ds^2}\right)^2 + 1 ds \geq \int_0^{2\pi} f^{-2} + v\left(\frac{df}{ds}\right)^2 ds.$$ Then we can take say $f = 1 + 0.5\cos s$. This ensures that the integral of $(df/ds)^2$ and $(d^2 f / ds^2)^2$ are the same, so the inequality becomes $$\int_0^{2\pi} \frac{1}{v}f^6 -2f^{-3}\frac{d^2 f}{ds^2} + 1 ds \geq \int_0^{2\pi} f^{-2} ds.$$ You can check that $$ \int_0^{2\pi} f^{-3}\frac{d^2 f}{ds^2} ds = 3\int_0^{2\pi} f^{-4} \left(\frac{d f}{ds}\right)^2 ds \geq 0.$$ On the other hand, by C-S we have $$\int_0^{2\pi} f^{-2} ds > 2\pi.$$ So when $v \to \infty$, the inequality cannot hold.