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Consider the following function: $F(A_1,\dots,A_n,\lambda_1,\dots,\lambda_n)=\sum_{i=1}A_i\Big(\frac{1}{(n-2)^2}\sum_{k,l\neq i}(\lambda_k-\lambda_l)^2-\frac{2n}{n-2}\lambda_i^2\Big),$ where $A_i\geq 0$ and $G(A_1,\dots,A_n,\lambda_1,\dots,\lambda_n)=\sum_{i=1}^nA_i\lambda_i-\frac{2R}{n-1}=0$ and $\sum_{i=1}^n\lambda_i=R.$ Here $R$ is a positive constant.

I want to prove that the function $F$ has a minimum. Here is my "proof" using Lagrange Multiplier:

Consider $F-\lambda G$. For $1\leq i\leq n$, we get $0=F_{A_i}-\lambda G_{A_i}=\frac{1}{(n-2)^2}\sum_{k,l\neq i}(\lambda_k-\lambda_l)^2-\frac{2n}{n-2}\lambda_i^2-\lambda\lambda_i.$ Substitute these equations into $F$, we get $F=\lambda\sum_{i=1}^nA_i\lambda_i=\frac{2\lambda R}{n-1}.$ On the other hand, we have $0=\sum_{i=1}^n(F_{A_i}-\lambda G_{A_i}),$ which implies that $0=\frac{n-1}{(n-2)^2}\sum_{k,l=1}^n(\lambda_k-\lambda_l)^2-\frac{2n}{n-2}\sum_{i=1}^n\lambda_i^2-\lambda R $ or equivalently, $\lambda R=\frac{n-1}{(n-2)^2}\sum_{k,l=1}^n(\lambda_k-\lambda_l)^2-\frac{2n}{n-2}\sum_{i=1}^n\lambda_i^2.$ The function on the right hand side of the last expression attains minimum when $\lambda_1=...=\lambda_n=\frac{R}{n}$. This implies that $\lambda R\geq -\frac{2R^2}{n-2}.$ Substituting back to the equation above, we obtain $F=\frac{2\lambda R}{n-1}\geq-\frac{4R^2}{(n-1)(n-2)}.$

My question is: Is my "proof" is correct? Since I feel there are something wrong in my calculation, I would be appreciated if someone can point out the mistakes in my "proof".

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    @D$i$dier Piau: Oh, I am sorry for making the mistake.2011-06-13

1 Answers 1

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Before starting with Lagrange multipliers one should simplify the occurring expressions somewhat. I propose the following:

Let $\sum_{i=1}^n \lambda_i^2=:S$. Then $\sum_{k,l}(\lambda_k-\lambda_l)^2=\sum_{k,l}(\lambda_k^2-2\lambda_k\lambda_l+\lambda_l^2)=2n S-2 R^2$ and therefore (using this formula for $n-1$ in place of $n$) $\sum_{k,l\ne i}(\lambda_k-\lambda_l)^2={n-1\over n}(S-\lambda_i^2)-2(R^2-2R\lambda_i+\lambda_i^2)\ .$ As $\sum_i A_i \lambda_i$ has a prescribed value R' you are left with a problem of the form \sum_i p_i \lambda_i^2=\min\ , \qquad \sum_i \lambda_i=R\ , \quad \sum_i A_i \lambda_i=R'\ , \qquad \sum_{i=1}^n \lambda_i^2=S.