Edit: question has been changed from 'lexicographic' (cf. "D K"'s answer below) to 'degree' minimality.


Let $x\_1,x\_2,x\_3$ be indeterminates. Fix an integer $k> 3$. Consider the set $M$ of all monomials of the form $x\_1^{i\_1}.x\_2^{i\_2}.x\_3^{i\_3}$ where each $i\_j\in \mathbb{N}$ with $i\_j\geq 1$ and $i\_1+i\_2+i\_3 = k$.


Fix an arbitrary $F\in M$ and write $F = x\_1^{a}.x\_2^{b}.x\_3^{c}$.


**Question** When does there exist a triple $(w\_1, w\_2, w\_3)$, with each $w\_i\in \mathbb{N}$ and $w\_i\geq 1$, such that $$a.w\_1 + b.w\_2 + c.w\_3 < i\_1.w\_1 + i\_2.w\_2 + i\_3.w\_3, \forall G = x\_1^{i\_1}.x\_2^{i\_2}.x\_3^{i\_3}\in M \;\text{with} \; G\neq F ?$$ In the exponential case, same question replaced by $a^{w\_1}+ b^{w\_2}+ c^{w\_3} < i\_1^{w\_1}+i\_2^{w\_2}+ i\_3^{w\_3}$.


Positive example in additive case: $k = 4$ on three indeterminates. Then if $F$ corresponds to the element of $B\_4$ with exponents $(2,1,1)$, then for any $w\_2, w\_3>1$, $w = (1,w\_2,w\_3)$ has the desired property.


Sometimes this question has a negative answer: e.g. k=4 on only two indeterminates. if we take F to correspond to the element in $B\_4$ with exponents $(2,2)$ then we cannot achieve both $2w\_1 + 2w\_2 < w\_1 + 3 w\_2$ and $2w\_1+2w\_2 < 3w\_1+w\_2$.




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Same question for $n$-variables $x\_1,\ldots, x\_n$ and $k > n.$


Apologies if this question has been answered; (general and specific) references also appreciated. 



Question on weights and minimal degree 
I haven't completely checked, but examples suggest that $F$ must be, up to reordering the $x\_i$'s, the minimal lexicographic element in $B\_k$ (in general, you can also replace $B\_k$ with a subset of $B\_k$, and then take the minimal element of this new set)


Example: take $k=5$. Then $(a,b,c)$ must have one of $a,b,$ or $c$ equal to $3$, i.e. no exponent of 2 is allowed in $F$ for a positive answer. 


Explicitly, for the affirmative: we can assume without loss of generality that $F = xyz^3$. Then we get that $w\_3 < w\_1$, $w\_3 < w\_2$ and $2w\_3 < w\_1+w\_2$, which is infinitely possible. Take, e.g. $w = (2,2,1)$. On the other hand if $F = xy^2z^2$ there is no solution since we must satisfy $w\_3 < w\_2$ and $w\_2 < w\_3$.


