The 36th KOREAN MATHEMATICAL OLYMPIAD Final Round

It suffices to prove the following. $$\sum _{i=1}^n{a}_i(b_i-b_{i+1})\le \frac{n}{8}\sum _{i=1}^n\left[(a_i-a_{i+1}{)}^2+(b_i-b_{i+1}{)}^2\right]$$ By replacing $b_i$ by $b_{i+j}$ in the above equation and adding up for $j=0,1,2$, we can obtain our desired result. Let $$\mathcal{R}=\frac{1}{2}\sum _{i=1}^n\left[(a_i-a_{i+1}{)}^2+(b_i-b_{i+1}{)}^2\right]$$ and consider $$S_j=\sum _{i=1}^n{a}_{i+j}(b_i-b_{i+1})$$ where $j$ is integer. (We consider any indices as modulo $n$, so that $a_{i+nk}=a_i$ holds for every integers $i,k$.) We can observe $$\begin{array}{rl}|S_j-S_{j+1}|& =\left|\sum _{i=1}^n(a_{i+j}-a_{i+j+1})(b_i-b_{i+1})\right|\\ & \le \frac{1}{2}\sum _{i=1}^n\left((a_{i+j}-a_{i+j+1}{)}^2+(b_i-b_{i+1}{)}^2\right)\\ & =\frac{1}{2}\left(\sum _{i=1}^n(a_{i+j}-a_{i+j+1}{)}^2+\sum _{i=1}^n(b_i-b_{i+1}{)}^2\right)\\ & =\mathcal{R}\end{array}$$ and obtain the following as its result: $$|S_0-S_j|\le |j|\mathcal{R}.$$ Meanwhile we have $$\sum _{j=0}^{n-1}{S}_j=\sum _{i=1}^n\left(\sum _{j=0}^{n-1}{a}_{i+j}\right)(b_i-b_{i+1})=\left(\sum _{j=0}^{n-1}{a}_j\right)\left(\sum _{i=1}^n(b_i-b_{i+1})\right)=0$$ so for any $-n<k<n$ we have $$n{S}_0=\sum _{j=-k+1}^{n-k}(S_0-S_j)\le \sum _{j=-k+1}^{n-k}|S_0-S_j|\le \mathcal{R}\sum _{j=-k+1}^{n-k}|j|.$$ If $n$ is even then we let $k=n/2$ to obtain $$n{S}_0\le \mathcal{R}\sum _{j=-n/2+1}^{n/2}|j|=\frac{n^2}{4}\mathcal{R}$$ and if $n$ is odd then we let $k=(n+1)/2$ to obtain $$n{S}_0\le \mathcal{R}\sum _{j=-(n-1)/2}^{(n-1)/2}|j|=\frac{n^2-1}{4}\mathcal{R}<\frac{n^2}{4}\mathcal{R}.$$ In any cases, we have $$S_0\le \frac{n}{4}\mathcal{R}=\frac{n}{8}\sum _{i=1}^n\left[(a_i-a_{i+1}{)}^2+(b_i-b_{i+1}{)}^2\right]$$ thus proving our inequality. $\square$

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Alternative solution.

We will show the following inequality as in the Solution 1. $$\sum _{i=1}^n{a}_i(b_i-b_{i+1})\le \frac{n}{8}\sum _{i=1}^n\left[(a_i-a_{i+1}{)}^2+(b_i-b_{i+1}{)}^2\right]$$ Let $\bar{a}$ be the average of all $a_i$. Then above is equivalent to: $$\sum _{i=1}^n(a_i-\bar{a})(b_i-b_{i+1})\le \frac{n}{8}\sum _{i=1}^n\left[(a_i-a_{i+1}{)}^2+(b_i-b_{i+1}{)}^2\right].$$ By noting the following (follows from AM-GM) $$\sum _{i=1}^n(a_i-\bar{a})(b_i-b_{i+1})\le \frac{2}{n}\sum _{i=1}^n(a_i-\bar{a}{)}^2+\frac{n}{8}\sum _{i=1}^n(b_i-b_{i+1}{)}^2$$ it suffices to prove $$\sum _{i=1}^n(a_i-\bar{a}{)}^2\le \frac{n^2}{16}\sum _{i=1}^n(a_i-a_{i+1}{)}^2.$$ We let $$M=\max a_i-\min a_i$$ and we will obtain bounds for both sides using $M$. Lemma 4. We have $$\sum _{i=1}^n(a_i-\bar{a}{)}^2\le \frac{n}{4}{M}^2.$$ Proof. Both sides of the equation are invariant under adding same constant to all $a_i$, so it suffices to show when $(\max a_i,\min a_i)=(M/2,-M/2)$. In that case, we can show: $$\sum _{i=1}^n(a_i-\bar{a}{)}^2=\sum _{i=1}^n{a}_i^2-n{\bar{a}}^2\le \sum _{i=1}^n{a}_i^2\le \sum _{i=1}^n(M/2{)}^2=\frac{n}{4}{M}^2.$$ Lemma 5. We have $$\sum _{i=1}^n(a_i-a_{i+1}{)}^2\ge \frac{4}{n}{M}^2.$$ Proof. Let $a_i,a_j$ be the maximum and minimum among $a_1,\dots ,a_n$ respectively, and assume $i<j$ without loss of generality. Using the Cauchy-Schwarz inequality we have $$M^2={\left(\sum _{l=i}^{j-1}(a_l-a_{l+1})\right)}^2\le (j-i)\sum _{l=i}^{j-1}(a_l-a_{l+1}{)}^2$$ and similarly $$M^2={\left(\sum _{l=j}^{n+i-1}(a_l-a_{l+1})\right)}^2\le (n+i-j)\sum _{l=j}^{n+i-1}(a_l-a_{l+1}{)}^2.$$ Thus we have (the last part uses AM-HM) $$\begin{array}{rl}\sum _{l=1}^n(a_l-a_{l+1}{)}^2& =\sum _{l=i}^{j-1}(a_l-a_{l+1}{)}^2+\sum _{l=j}^{n+i-1}(a_l-a_{l+1}{)}^2\\ & \le \frac{M^2}{j-i}+\frac{M^2}{n+i-j}\\ & \le \frac{4M^2}{n}.\end{array}$$ Combining those two lemmas yield the desired result of $$\sum _{i=1}^n(a_i-\bar{a}{)}^2\le \frac{n{M}^2}{4}\le \frac{n^2}{16}\sum _{i=1}^n(a_i-a_{i+1}{)}^2.$$

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Alternative solution.

We note that if $n=3$ then the left hand side (of our original inequality) becomes zero so our problem holds obviously. In this solution, we will prove the following inequality of the Solution 1 for $n\ge 4$. $$\sum _{i=1}^n{a}_i(b_i-b_{i+1})\le \frac{n}{8}\sum _{i=1}^n[(a_i-a_{i+1}{)}^2+(b_i-b_{i+1}{)}^2]$$ We will consider sum of the following two inequalities. $$\begin{array}{rl}\sum _{i=1}^n(a_i-a_{i+1})(b_i-b_{i+1})& \le \frac{1}{2}\sum _{i=1}^n[(a_i-a_{i+1}{)}^2+(b_i-b_{i+1}{)}^2]\\ \sum _{i=1}^n(a_i+a_{i+1})(b_i-b_{i+1})& \le \frac{\cot (\pi /n)}{2}\sum _{i=1}^n[(a_i-a_{i+1}{)}^2+(b_i-b_{i+1}{)}^2]\end{array}$$ The first one follows easily from AM-GM. For the second one, we consider a $n$-gon whose vertices have coordinates $(a_i,b_i)$. Then we can interpret its the left hand and righthand sides as two times its (signed) area and the sum of squares of its sides respectively. By considering the isoperimetric inequality for $n$-gon and Cauchy-Schwarz inequality, one can show their ratio is maximized for regular $n$-gon, so it suffices to check equality holds for regular $n$-gon case. Summing those two gives $$\sum _{i=1}^{n}2a_i(b_i-b_{i+1})\le \left(\frac{1}{2}+\frac{\cot (\pi /n)}{2}\right)\sum _{i=1}^n\left[(a_i-a_{i+1}{)}^2+(b_i-b_{i+1}{)}^2\right]$$ so it suffices to show $$\frac{1}{2}+\frac{\cot (\pi /n)}{2}\le \frac{n}{4}$$ for $n\ge 4$. When $n\ge 6$, we use $\cot (x)=(\tan (x){)}^{-1}<1/x$ and $\pi >3$ to show $$\frac{1}{2}+\frac{n}{2\pi }<\frac{1}{2}+\frac{n}{6}\le \frac{n}{4}.$$ For $n=4$ and $n=5$, we can prove it by explicitly calculating $\cot (\pi /n)$. ($\cot (\pi /4)=1$, $\cot (\pi /5)=\sqrt{1+\frac{2}{\sqrt{5}}}$)

Remark. The 'optimal constant' for this inequality can be given as $$C_{op}=\frac{(1+2\cos (2\pi /n))}{4\sin (\pi /n)}$$ instead of $3n/8$. Consider a vector space $V=\{(x_1,\dots ,x_n):\sum x_i=0\}$ and an operator $T$ on $V$ defined as $T((x_i))=(x_{i+1})$. Then our inequality can be expressed as follows. (The absolute value denotes the ordinary Euclidean length induced from $V\le {\mathbb{R}}^n$) $$⟨a,(1-T^3)b⟩\le C(|(1-T)a{|}^2+|(1-T)b{|}^2)$$ The operator $T$ on $V$ is orthogonal, and it can be diagonalized by complex orthogonal basis $v_k=({\zeta }_n^{kj}{)}_{j=1,\dots ,n}$ ($1\le k<n$) as $T{v}_k={\zeta }_n^k{v}_k$ (${\zeta }_n=\exp (2\pi i/n)$). Thus $1-T$ is invertible, and we can express the above inequality as follows. ($u=(1-T)a,v=(1-T)b$) $$⟨(1-T{)}^{-1}u,(1+T+T^2)v⟩=⟨u,(1-T{)}^{-1}(1+T+T^2)v⟩\le C(|u{|}^2+|v{|}^2)$$ One can see that $C$ can be given as the operator norm of $S=(1-T{)}^{-1}(1+T+T^2)$. As $S$ is normal operator, its operator norm is given as maximum of absolute value of its eigenvalues $|(1-{\zeta }_n^k{)}^{-1}(1+{\zeta }_n^k+{\zeta }_n^{2k})|$. One can observe that this obtains maximum $C_{op}$ when $k=1$ or $k=n-1$. $\square$