An Analytic Inequality

Question

$$\frac 12\ln(AC-B^2)\leq\frac{1}{2\pi}\int_{0}^{2\pi}\ln(A\sin^2\theta+2B\sin\theta\cos\theta+C\cos^2\theta)d\theta(A,C>0)$$ I see this inequality from a book about Honeycomb structure.It can be proved by dividing $[0,2\pi]$. $$|AC-B^2|^{\frac n2}\leq\prod_{l=1}^{n}\left[A\sin^2\frac{\pi(2l-1)}{n}-2B\sin\frac{\pi(2l-1)}{n}\cos\frac{\pi(2l-1)}{n}+C\cos^2\frac{\pi(2l-1)}{n}\right]$$ but I think there must be a direct way to prove this inequality.Any help will be thanked.

Answer 1

The integral can be evaluated in closed form. The expression under the logarithm is $$A\frac{1-\cos 2\theta}{2}+B\sin 2\theta+C\frac{1+\cos 2\theta}{2}=\frac{A+C}{2}+\frac{C-A}{2}\cos 2\theta+B\sin 2\theta,$$ which is equal to $A'+B'\cos(2\theta-\phi)$ with $$A'=\frac{A+C}{2},\qquad B'=\sqrt{B^2+\Big(\frac{A-C}{2}\Big)^2},$$ and some value of $\phi$ we don't care about, since if $f$ is $2\pi$-periodic, we have $$\int_0^{2\pi}f(2\theta-\phi)~d\theta=\int_0^{2\pi}f(2\theta)~d\theta=\int_0^{2\pi}f(\theta)~d\theta.$$ Thus we're left with $\int_0^{2\pi}\ln(A'+B'\cos\theta)~d\theta$, which can be evaluated using $$\int_0^{2\pi}\ln(1+d\cos\theta)~d\theta=2\pi\ln\frac{1+\sqrt{1-d^2}}{2}\qquad(|d|<1)$$ where the principal values are taken (see this answer of mine); the result is $$2\pi\ln\color{lightgray}{\left[\color{black}{\frac{1}{2}\left(\frac{A+C}{2}+\sqrt{AC-B^2}\right)}\right]}.$$