# How to prove this inequality of Karamata type?

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.

As an example of Question 2, consider $$f(x)=\ln{\sin{x}}$$, $$0. Since $$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}$$

• It is certainly not that general. The last inequality may fail for a general concave function. May 29, 2019 at 1:18
• The questions arise: what for? where could it be applied? May 29, 2019 at 7:33
• math.stackexchange.com is a right forum for such type questions. May 29, 2019 at 7:42
• did you check the inequality $f(x+y)-\frac12(f(x)+f(y))+f(\pi/n)-f(2\pi/n)\geqslant c(x+y-2\pi/n)$ for $c=f'(2\pi/n)-\frac12 f'(\pi/n)$? May 29, 2019 at 9:38
• "and the numerator grows" Why? Jun 21, 2019 at 2:44

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 $$k=2$$.

• Apparently, there are infinitely many different points where the inequality is achieved, at least in the case for $n = 4$ and $n=5.$ I am kind of failing to see how these points will be obtained from your solution. Jul 3, 2019 at 21:05
• @dezdichado For $n=4$ it is easy (see my response to Fedor Petrov in the comments to the question). $n=5$ is a bit trickier to understand. Jul 3, 2019 at 23:34
• I actually worked out the configuration for $n = 5,$ but exhausting all of the equilibrium points was rather tedious. Jul 4, 2019 at 19:43
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.