In arbitrary cyclic polygon then $\sum_{i<j} x_{ij}^\alpha \ge \sum_{i<j} y_{ij}^\alpha $

Question

I am looking for a proof of the inequality as follows:

Conjecture: Let $A_1A_2....A_n$ be the regular polygon incribed a circle $(O)$. Let $B_1B_2....B_n$ be a polygon incribed the circle $(O)$. Denote $x_{ij}=A_iA_j$ and $y_{ij}=B_{i}B_{j}$ then:

$$\sum_{i<j} x_{ij}^\alpha \ge \sum_{i<j} y_{ij}^\alpha $$

Where $ 1 \leq \alpha \leq 2$.

• The case $ \alpha = 1 $ was proved in our paper in here

• The case $ \alpha = 2 $ was proved in here

Example:

• $n=3$, let $ABC$ be a triangle with sidelength $a, b, c$ then we have the inequality as follows:

$$a^\alpha + b^\alpha+ c^\alpha \leq 3\times 3^{\frac{\alpha}{2}}R^\alpha$$ Where $ 1 \leq \alpha \leq 2$, $R$ is circumradius.

See also:

Answer 1

Here is a classic way of proving this inequality for a triangle. We fix our circle $(C)$ and take the triangle $ABC$ inscribed in $(C)$. Fix a side say $BC$ we want to prove that moving the point $A$ along its position on arc $\overparen{BC}=a$ until reaching the arc midpoint call it the Mount (the sides become equal) increases the quantity $b^{\alpha}+c^{\alpha}$ for $\alpha\le 2$. In fact we prove it directly for $\alpha=2$ then a functional argument gives the general case.

The quantity $b^{\alpha}+c^{\alpha}+a^\alpha$ of the left side is necessarily maximal when all sides are equal and in that case the side length is $\sqrt{3}R$.

For $\alpha=2$, notice that at the Mount the area of triangle $ABC$ increases as $bc \sin(\hat{A})$ increases so that $bc$ increases. Also as $b^2+c^2-2bc\cos(\hat{A})=cte$, assuming $\hat{A}$ is acute, $b^2+c^2$ increases. This reasoning applied successively tends towards an equilateral triangle.

For the functional argument write when moving to the Mount $b^{\alpha}=f^{\alpha}(\beta)$ and $c^{\alpha}=g^{\alpha}(\beta)$ for some regular functions $f,g$; where $\beta=\widehat{ABC}$. Assume without loss of generality that $f(\beta)\le g(\beta)$ so $f'(\beta)\ge 0$ we know that

$f'(\beta)f(\beta)+g'(\beta)g(\beta)\ge 0$ or $f'(\beta)\dfrac{f(\beta)}{g(\beta)}+g'(\beta)\ge 0$

So necessarily $f'(\beta)\dfrac{f^{\alpha-1}(\beta)}{g^{\alpha-1}(\beta)} +g'(\beta)\ge 0$ when $\alpha\le 2$.

Answer 2

Here goes a general conditional claim, which gives a full proof for $n=4$. Possibly this may be extended to other values of $n$.

Definition. Let $h(z)$ be a function from the unit circle $\mathbb{S}$ to $\mathbb{R}$. We call $h$ $n$-admissible, if there exist real coefficients $c_0$ and non-negative $c_m$, where $m$ runs over positive integers non-divisible by $n$, such that the inequality $$h(z)\leqslant c_0-\sum_m c_m(z^m+z^{-m}) \tag{1}$$ holds for all $z\in \mathbb{S}$, and equality holds in (1) whenever $z^n=1$ and $z\ne 1$.

Claim. For an $n$-admissible function $h\colon \mathbb{S}\to \mathbb{R}$ the energy functional $F(y_1,\ldots,y_n):=\sum_{1\leqslant k<j\leqslant n} h(y_k/y_j)$ for $y_1,\ldots,y_n\in \mathbb{S}$ takes the maximal value when $y$'s are the vertices of the regular $n$-gon.

Proof. By (1) we have $$F(y_1,\ldots,y_n)\leqslant {n\choose 2}c_0-\sum_mc_m\sum_{k<j}(y_k^m/y_j^m+y_j^m/y_k^m),\tag{2}$$ and by equality cases (2) is an equality for a regular polygon. Thus, it suffices to check that for every $m$ not divisible by $n$ the sum $\sum_{k<j}(y_k^m/y_j^m+y_j^m/y_k^m)$ takes minimal value for a regular $n$-gon. Indeed, $$\sum_{k<j}(y_k^m/y_j^m+y_j^m/y_k^m)=-n+|\sum_{k}y_k^m|^2\geqslant -n$$ with equality iff $\sum_k y_k^m=0$, which is true for a regular $n$-gon since $n$ does not divide $m$.

So, your conjecture for $n=4$ follows from Claim and the following

Proposition. If $\alpha=2\beta$ for $0\leqslant \beta\leqslant 1$, then $h(z):=|1-z|^\alpha$ is a 4-admissible function.

Proof. We have $h(z)=|2-z-1/z|^\beta$. Denoting $z=e^{i\theta}$ and $t=\cos \theta$, we have $z^2+z^{-2}=2\cos 2\theta=2t^2-1$, and (1) would follow from the bound $$(2-2t)^\beta\leqslant c_0-2c_1 t-2c_2(2t^2-1)\tag{3}$$ for $t\in [-1,1]$ which turns into equality for $t=0$ and $t=-1$. I claim that (3) holds with $c_0=2^{\beta-1}(1-\beta+2^\beta)$, $2c_1=\beta2^{\beta}$, $2c_2=2^{\beta-1}(1+\beta-2^\beta)$. We should check several things.

1. $c_1,c_2$ are non-negative. This is clear for $c_1$ and for $c_2$ this follows from $2^x$ being a convex function whose graph on $[0,1]$ lies below the chord $y=x+1$ joining the points $(0,1)$ and $(1,2)$.

2. Equality cases in (3) for $t=0$ and $t=1$ are straightforward.

3. At the point $t=0$, the derivatives of both sides of (3) are equal to $-2^\beta$.

4. (3) holds at the point $t=1$. Indeed, this reads as $0\leqslant c_0-2c_1-2c_2=-\beta 2^{\beta+1}+4^\beta=2^{\beta+1}(2^{\beta-1}-\beta)$. We have $2^{\beta-1}\geqslant e^{\beta-1}\geqslant 1+(\beta-1)=\beta$ due to the useful inequality $e^x\geqslant 1+x$, which holds for all real $x$ by convexity of $e^x$ or other reasons.)

5. That's enough. Indeed, if we denote $A(t):=-(2-2t)^\beta+c_0-2c_1 t-2c_2(2t^2-1)$, then we already know that $A$ has a root at $t=-1$, a double root at $t=0$, and is non-negative at $t=1$. If it is non non-negative on the whole segment $[-1,1]$, it must have another root (or 0 is a root of multiplicity at least 3). So, totally at least 4 roots of $A$ on $[-1,1]$. Then $A'''$ must have a root on $(-1,1)$ by Rolle. But it does not (unless $\beta=0$ or $\beta=1$ when $A$ is identical 0).