How to prove this inequality of Karamata type?

Question

Question 1:

Let $x_{i}>0$, ($i=1,2,\cdots,n$) and such that $$x_{1}+x_{2}+\cdots+x_{n}=\pi.$$ Show that $$ \dfrac{\sin{x_{1}}\sin{x_{2}}\cdots\sin{x_{n}}}{\sin{(x_{1}+x_{2})}\sin{(x_{2}+x_{3})}\cdots\sin{(x_{n}+x_{1})}}\le\left(\dfrac{\sin{\frac{\pi}{n}}}{\sin{\frac{2\pi}{n}}}\right)^n $$ Question 2 (may not hold): if $f''(x)\le 0,x\in I$, can we prove the following inequality? $$ \begin{split} f(x_{1}+x_{2})+&f(x_{2}+x_{3})+\ldots+f(x_{n}+x_{1})+nf\left(\dfrac{x_{1}+x_{2}+\ldots+x_{n}}{n}\right)\\ &\ge f(x_{1})+f(x_{2})+\ldots+f(x_{n})+nf\left(\dfrac{2(x_{1}+x_{2}+\ldots+x_{n})}{n}\right), \end{split} $$ where $x_{i}\in I$, $i=1,2,3\ldots,n$. I tried everything, but failed.

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As an example of Question 2, consider $f(x)=\ln{\sin{x}}$, $0<x<\pi$. Since $$f''(x)=-\csc^2{x}<0$$ it suffices to prove that $$ \begin{split} f(x_{1}+x_{2})+f(x_{2}+x_{3})+&\ldots+f(x_{n}+x_{1})+nf\Big(\dfrac{\pi}{n}\Big)\\ &\ge f(x_{1})+f(x_{2})+\ldots+f(x_{n})+nf\Big(\dfrac{2\pi}{n}\Big) \end{split} $$ or $$ \begin{split} f(x_{1}+x_{2})+&f(x_{2}+x_{3})+\ldots+f(x_{n}+x_{1})+nf\left(\dfrac{x_{1}+x_{2}+\ldots+x_{n}}{n}\right)\\ &\ge f(x_{1})+f(x_{2})+\ldots+f(x_{n})+nf\left(\dfrac{2(x_{1}+x_{2}+\ldots+x_{n})}{n}\right). \end{split}$$ In other words,if $f''(x)\le 0$, can we prove following inequality? $$ \begin{split} f(x_{1}+x_{2})+&f(x_{2}+x_{3})+\ldots+f(x_{n}+x_{1})+nf\left(\dfrac{x_{1}+x_{2}+\ldots+x_{n}}{n}\right)\\ &\ge f(x_{1})+f(x_{2})+\ldots+f(x_{n})+nf\left(\dfrac{2(x_{1}+x_{2}+\ldots+x_{n})}{n}\right)? \end{split} $$

Answer 1

It is actually quite a cute problem except it is only 1/3-analysis and 2/3 elementary geometry. The analysis part is that if you have $2$ positive numbers $A,B$, then for any $p,q$ with $\frac 1p+\frac 1q=1$, we have $A+B=\frac{pA}{p}+\frac{qB}q\ge p^{1/p}q^{1/q}A^{1/p}B^{1/q}$ (Young) and if $pA=qB$, you have equality.

The geometry part is Ptolemy's theorem.

Now let $d,d'$ be some consecutive diagonals of order $k\in[2,n-2]$ (sides are order $1$ (or $n-1$), diagonals spanning $2$ sides are order $2$ (or $n-2$, etc.) in an inscribed $n$-gon. Let $a,b$ be the diagonals of orders $k-1,k+1$ and $c,c'$ be the sides so that $d,d'$ are the diagonals of the quadrilateral with the sides $a,c,b,c'$. We have $$ dd'=ab+cc'\ge p^{1/p}q^{1/q}(ab)^{1/p}(cc')^{1/q} $$ and, most importantly, we can choose $p,q$ depending on $k$ only so that we have an identity for the regular $n$-gon. Multiplying over all choices of the pair $d.d'$, we get an inequality of the kind $$ {\prod}_k\ge c_k\left[{\prod}_{k-1}{\prod}_{k+1}\right]^{\alpha_k}{\prod}_1^{1-2\alpha_k} $$ where $\prod_k$ is the product of diagonals of order $k$, $c_k>0$ and $\alpha_k\in(0,\frac 12)$ are some appropriate numbers.

The rest is trivial. If you want more analytic flavor, just iterate this inequality like crazy until everything except $\prod_1$ wears out on the RHS, but you can also do it by completely elementary means (you have some fancy kind of log-concavity here). The upshot is that you get an inequality $\prod_k\ge C_k\prod_1$ in which the regular $n$-gon produces an identity. Now just take $k=2$.

I'm answering here on MO because the MSE thread is cluttered enough already, but your best bet for asking such stuff is, probably, AoPS.

Answer 2

The answer to your Question 2 is negative. E.g., let $f(x)=-\max(0,x-1)$ (or a smooth concave approximation to $f(x)$), $n=3$, $x_1=1/2,x_2=0,x_3=1$. Then your inequality becomes $-1/2\ge0$, which is false.