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".