MathLabs

Problem 5

Let nn be a positive integer and let x1,x2,…,xnx_1,x_2,\ldots,x_n be real numbers. Prove that (∑i=1n∑j=1n∣xi−xj∣)2≤2(n2−1)3∑i=1n∑j=1n(xi−xj)2\left(\sum_{i=1}^n\sum_{j=1}^n|x_i-x_j|\right)^2\le\frac{2(n^2-1)}3\sum_{i=1}^n\sum_{j=1}^n(x_i-x_j)^2, with equality if and only if x1,x2,…,xnx_1,x_2,\ldots,x_n form an arithmetic sequence.
Step 3 of 3: Centering and Cauchy–Schwarz
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In plain words

Center the variables so the equality condition is visible.

3L2≤(n2−1)Q3L^2\le(n^2-1)Q
Detailed analysis

Put xˉ=1n∑k=1nxk\bar x=\frac1n\sum_{k=1}^n x_k. Counting the coefficient of each xkx_k in the sum of all pairwise differences gives L=∑k=1n(2k−n−1)xk=∑k=1n(2k−n−1)(xk−xˉ)L=\sum_{k=1}^n(2k-n-1)x_k=\sum_{k=1}^n(2k-n-1)(x_k-\bar x), because ∑k=1n(2k−n−1)=0\sum_{k=1}^n(2k-n-1)=0. The finite-sum identities ∑k=1nk=n(n+1)2\sum_{k=1}^n k=\frac{n(n+1)}2 and ∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}6 give ∑k=1n(2k−n−1)2=4∑k=1nk2−4(n+1)∑k=1nk+n(n+1)2=n(n2−1)3\sum_{k=1}^n(2k-n-1)^2=4\sum_{k=1}^n k^2-4(n+1)\sum_{k=1}^n k+n(n+1)^2=\frac{n(n^2-1)}3. Expanding the square differences gives ∑i=1n∑j=1n(xi−xj)2=2n∑k=1nxk2−2(∑k=1nxk)2=2n∑k=1n(xk−xˉ)2\sum_{i=1}^n\sum_{j=1}^n(x_i-x_j)^2=2n\sum_{k=1}^n x_k^2-2(\sum_{k=1}^n x_k)^2=2n\sum_{k=1}^n(x_k-\bar x)^2, so Q=12∑i=1n∑j=1n(xi−xj)2=n∑k=1n(xk−xˉ)2Q=\frac12\sum_{i=1}^n\sum_{j=1}^n(x_i-x_j)^2=n\sum_{k=1}^n(x_k-\bar x)^2. Cauchy–Schwarz therefore yields L2≤n(n2−1)3∑k=1n(xk−xˉ)2=n2−13QL^2\le\frac{n(n^2-1)}3\sum_{k=1}^n(x_k-\bar x)^2=\frac{n^2-1}{3}Q. Equality holds exactly when xk−xˉ=c(2k−n−1)x_k-\bar x=c(2k-n-1) for one constant cc, i.e. xk=x1+2c(k−1)x_k=x_1+2c(k-1); after the relabeling this is an arithmetic sequence, including c=0c=0.