Let positive integers $a,b,c$ be pairwise coprime. Denote by $g(a, b, c)$ the maximum integer not representable in the form $xa+yb+zc$ with positive integral $x,y,z$. Prove that $$ g(a, b, c)\ge \sqrt{3abc}$$ I would like a proof of this fact. I was assured it is true. But I have no proof for this. I see this inequality is strong since $g(3,3k+1,3k+2)=9k+5$ but this is probably trivial but no improvement from there. Some help would be appreciated. Thanks.
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1I forget how the proof goes, but it is in Davison, J. L. "On the Linear Diophantine Problem of Frobenius." J. Number Th. 48, 353-363, 1994. – deinst Jul 12 '14 at 11:55
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http://en.wikipedia.org/wiki/Coin_problem gives $g(a,b,c)\ge\sqrt{3abc}-a-b-c$ with a reference to M. Beck and S. Zacks (2004). "Refined upper bounds for the linear Diophantine problem of Frobenius". Adv. Appl. Math. 32 (3): 454–467. arXiv:math/0305420 – Gerry Myerson Jul 12 '14 at 12:40
1 Answers
Reducing to publicly available articles.
Let $f(a,b,c)$ be the largest integer not representable as a positive linear combination of $a,b,$ and $c$, ie as $za+xb+yc$ with $x,y,z\in\mathbb{N}$
Assume wlog that $a<b<c$.
Let $$S=\{(x,y):bx=cy\bmod a, 0 < x,y\le a\}$$ $$T_1=\{(x,y):bx<cy, 0 < x,y\le a\}$$ $$T_2=\{(x,y):bx>cy, 0 < x,y\le a\}$$
Then let $$x_1 = \min\{x:(x,y)\in S\cap T_1\}$$ $$y_2 = \min\{y:(x,y)\in S\cap T_2\}$$ and define $x_2,y_1$ so that $(x_1,y_1)\in S$ and $(x_2,y_2)\in S$ then let $x_3 = x_1-x_2$ and $y_3 = y_2-y_1$. Then $0<x_3,y_3$ and $$f(a,b,c) = \max\{b x_1 + c y_3, b x_3 + c y_2 \}$$ and $x_1 y_2 - x_2 y_1 = a$.
This is due to Brauer and Shockley (thankfully it is online and publicly available) it also implicit in stanwagon.com/public/FrobeniusByLatticePoints.pdf
First we show that if $x,y,z,w \ge 0$ and $xw + xy + zy =1$ then $x+y+\max(z,w)\ge\sqrt{3}$. This is obviously true if at least one of $x,y,z,w\ge 2$ and we can wlog consider $w\le z$. Let us find the minimum of $x+y+z$ on $0\le x,y\le 2$, $0\le w\le z\le 2$ and $xw + xy + zy =1$. Grind through the Lagrange multiplier stuff and find that the minimum occurs at $w=z$ and $$\min\{x+y+\max(z,w)\}=\sqrt{3}$$
Now if we let $\alpha_i = x_i \sqrt{b/ac}$ and $\beta_i = y_i \sqrt{c/ab}$. Now let $\epsilon_1 = \alpha_1 - \beta_1$ and $\epsilon_2 = \beta_2-\alpha_2$ then $\epsilon_1,\epsilon_2>0$ and $$\epsilon_1 \alpha_2 + \epsilon_1 \epsilon_2 + \epsilon_2 \beta_1 = 1$$ and $$\frac{f(a,b,c)}{\sqrt{abc}}=\epsilon_1+\epsilon_2+\max(\alpha_1,\beta_2),$$ and it follows that $f(a,b,c) \ge \sqrt{3abc}$.
Added:
Intuitively what is happening is this:
For each residue class $\mod a$ there is minimum representable value of the form $a + bx + cy$, and $f(a,b,c)$ is the largest of these values. If we look at the integers $(x,y)$ such that $bx+cy\equiv 0 \bmod a$ they form a lattice. This lattice has a basis $(x_1,y_1), (x_2,y_2)$ such that $x_1>0$, $y_1\le 0$, $x_2\le 0$, $y_2>0$, $x_3=x_1+x2>0$ and $y_3=y_1+y_2>0$ (this is not obvious). Now if
we have $bx+cy = d\bmod a$ with $x,y\in\mathbb{N}$ and $x>x_1$ then $x-x_1, y-y_1\in\mathbb{N}$ and $b(x-x_1)+c(y-y_1) = d\bmod a$. We can similarly shrink by subtracting $(x_2,y_2)$ if $y>y_2$ and by $(x_3,y_3)$ if $x>x_3$ and $y>y_3$. So we only need consider the L-shaped region defined as $x\le x_1$, $y\le y_2$ and ($x\le x_3$ or $y\le y_3$). (If $x_2$ or $y_1$ is 0 then the L-shape degenerates to a rectangle.) This region has area $a$ so contains just the minimum representable values $\mod a$. So to find the lower bound we need to find the minimum value of the maximum of $bx+cy$ over all such L shaped regions.

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