Let $a,b\in\mathbb{N}$ with $\gcd(a,b)=1.$ Let $c\in\mathbb{Z}$ with $c>ab-b-a.$ Then, the equation $ax+by=c$ has a solution $(x,y)\in\mathbb{Z^2}$ with $x,y≥0.$
Since $1=\gcd(a,b)|c,$ we know that the equation has a solution $(x_0,y_0)\in\mathbb{Z^2}.$ Moreover, since $\gcd(a,b)=1,$ every other solution to this diophantine equation $(x,y)$ satisfies
$$x=x_0+bt$$ $$y=y_0-at$$
Here, $t\in\mathbb{Z}.$
We want $x,y≥0.$ So, $t≥\frac{-x_0}{b},$ and $\frac{y_0}{a}≥t.$ Putting this together, we get that $t$ must satisfy
$$\frac{-x_0}{b}≤t≤\frac{y_0}{a}.$$
So, we want the interval $[\frac{-x_0}{b},\frac{y_0}{a}]$ to contain an integer. A sufficient condition for this is $\frac{y_0}{a}+\frac{x_0}{b}>1.$
I thought assuming $\frac{y_0}{a}+\frac{x_0}{b}≤1$ would give a contradiction, but that just gives $c≤ab$ upon multiplying by $ab.$ This is not a contradiction.
How to proceed from here?
