For a field $F$ let $p(F)=p$ if the characteristic of $F$ is a prime $p$, and $p(F)=+\infty$ if $F$ is of characteristic zero. 

In 2007 I considered the linear extension of the Erdos-Heilbronn conjecture, and conjectured that (cf. http://arXiv.org/abs/0810.0467) for any nonzero elements $\lambda_1,\ldots,\lambda_n$ of a field $F$ with $p(F)\not=n+1$ and a finite subset $A$ of $F$ we have
\begin{align}&|\{\lambda_1a_1+\ldots+\lambda_n a_n:\ a_1,\ldots,a_n\ \text{are distinct elements of}\ A\}|
\\&\qquad \qquad \ge\min\{p(F)-\delta,\, n(|A|-n)+1\},\end{align}
where $\delta$ is $1$  if $n=2$ and $\lambda_1+\lambda_2=0$, and $\delta=0$ otherwise.

Motivated by the above as well as [Question 316142][1] of mine, here I ask the following question.

QUESTION: Is my following conjecture true?

**Conjecture**. Let $\lambda_1,\ldots,\lambda_n\ (n\ge3)$ be positive integers with $\lambda_1\le\ldots\le\lambda_n\le n$ and $\gcd(\lambda_1,\ldots,\lambda_n)=1$. Let $F$ be a field with $p(F)>n+1$. Then, for any finite subset $A$ of $F$ with $|A|\ge n+\delta_{n,3}$ we have
\begin{align}&\bigg|\bigg\{\sum_{k=1}^n\lambda_ka_k:\ a_1,\ldots,a_n\ \text{are distinct elements of}\ A\bigg\}\bigg|
\\\ge&\min\bigg\{p(F),\ (\lambda_1+\ldots+\lambda_n)(|A|-n)+\sum_{k=1}^{\lfloor n/2\rfloor}(n+1-2k)(\lambda_{n+1-k}-\lambda_k)+1\bigg\}.\end{align}

Now let me explain where the lower bound comes from.  Suppose that $A$ is just the subset $\{1,\ldots,m\}$ of the rational field $\mathbb Q$. For the set
$$S=\{\lambda_1a_1+\ldots+\lambda_na_n:\ a_1,\ldots,a_n\ \text{are distinct elements of}\ A\},$$
its minimal element should be $\sum_{k=1}^n\lambda_k(n+1-k)$ while its maximal element should be $\sum_{k=1}^n\lambda_k(m-n+k)$. Note that
\begin{align}&\bigg|\bigg\{\sum_{k=1}^n\lambda_k(n+1-k),\ 
 \ldots,\ \sum_{k=1}^n\lambda_k(m-n+k)\bigg\}\bigg|
\\=&(\lambda_1+\ldots+\lambda_n)(m-n)+\sum_{k=1}^{\lfloor n/2\rfloor}(n+1-2k)(\lambda_{n+1-k}-\lambda_k)+1.
\end{align}
If $\lambda_k=k$ for all $k=1,\ldots,n$, then
$$\sum_{k=1}^{\lfloor n/2\rfloor}(n+1-2k)(\lambda_{n+1-k}-\lambda_k)=\sum_{k=1}^{\lfloor n/2\rfloor}(n+1-2k)^2=\frac{n(n^2-1)}6.$$

Any comments are welcome!

[1]: https://mathoverflow.net/questions/316142