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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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@Paul: (i) Do you want $\lambda_i \ge 0$? (ii) Are the $A_i$ fixed constants? (iii) If the answer to (i) is yes, then there is a minimum. So then you presumably want to find the minimum. (iv) This does not look like any Lagrange multiplier argument I have ever seen, where are the partial derivatives? (v) Maybe I can look at it, but not immediately, and not if I am not given the answers to (i) and (ii). –  André Nicolas Jun 12 '11 at 14:03
    
Related post: math.stackexchange.com/q/44707/6179 In such cases it is good practice to include a link to the other post. // Re (iv): $F_{A_i}$ and $G_{A_i}$ are most probably partial derivatives. // Re (i): I find fascinating that we still do not know whether the OP means the condition $\lambda_i\ge0$ to hold or not, given that this problem was already raised on the other page. –  Did Jun 12 '11 at 16:54
    
@Didier Piau: I wondered about $F_{A_i}$, expected partials with respect to the $\lambda_j$. –  André Nicolas Jun 12 '11 at 18:17
    
@user6312: Yes. –  Did Jun 12 '11 at 20:11
    
@Didier Piau: Re (i): No. We don't assume $\lambda_i\geq 0$. Re (ii): No, $A_i$ are not fixed. Instead they are variables. But they are positive. That's why I wrote $F(A_1,...A_n,\lambda_1,...\lambda_n)$. Re (iv): The partial derivatives are $F_{A_i}-\lambda G_{A_i}=\frac{\partial F}{\partial A_i}-\lambda\frac{\partial G}{\partial A_i}$. –  Paul Jun 12 '11 at 22:29
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1 Answer

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 $$\sum_i p_i \lambda_i^2=\min\ , \qquad \sum_i \lambda_i=R\ , \quad \sum_i A_i \lambda_i=R'\ , \qquad \sum_{i=1}^n \lambda_i^2=S.$$

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