<s>[INST] Proof of inequality without calculus 
So we are given the equation $3x+4y+xy=2012$ where $x$ and $y$ are positive integers. Prove that $x+y\geq83$.

Using calculus optimisation methods, this can be proved. However, it requires a lot of difficult calculations, and students are not expected to know calculus to answer this. How else could you prove this? You could always go through each possible value, but that would take a very long time.
 [/INST] You can rearrange the equation $3x+4y+xy=2012$:
\begin{align}
3x+4y+xy\quad =\quad 2012\\ 3x+4y+xy+12\quad =\quad 2024\\ x(y+3)+4y+12\quad =\quad 2024\\ x(y+3)+4(y+3)\quad =\quad 2024\\ (y+3)(x+4)\quad =\quad 2024\\ 
\end{align}
And from the AM-GM inequality:
$\frac{a+b}{2}\ge\sqrt{ab}$
And using $a = x+4$ and $b = y+3$ we get:
\begin{align}
\frac{x+4+y+3}{2}&\ge\sqrt{(x+4)(y+3)}\\
\frac{x+y+7}{2}&\ge\sqrt{2024}\\
x+y+7&\ge2\sqrt{2024}\\
x+y&\ge2\sqrt{2024}-7\\
x+y&\ge83
\end{align}
Note:
\begin{align}
{(2\sqrt{2024})}^{2}-{89}^{2} &= 8096-7921>0 \\
2\sqrt{2024}&>89 \\
2\sqrt{2024}-7&>82 \\
x+y&>82\\
x+y&\ge 83 \text{ (because x, y are integers) }
\end{align}
In general we have for $a,b,c\in \mathbb{{C}^{*}}$:
\begin{align}
ax+by+cxy&=x(a+cy)+by\\&=cx\left( y+\frac { a }{ c }  \right) +by\\&=cx\left( y+\frac { a }{ c }  \right) +by+\frac { ab }{ c } -\frac { ab }{ c } \\ &=cx\left( y+\frac { a }{ c }  \right) +b\left( y+\frac { a }{ c }  \right) -\frac { ab }{ c } \\&=\left( y+\frac { a }{ c }  \right) \left( b+cx \right) -\frac { ab }{ c } 
\end{align}
</s>