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.
 A: We have $$xy +3x +4y +12 =2024$$ thus $$(x+4)(y+3) =2024$$
hence $$\left(\frac{x+y+7}{2}\right)^2 \geq (x+4)(y+3) =2024 >44.5^2$$
therefore $$\frac{x+y+7}{2} \geq 45 $$
A: Since we have 
$$3x+4y+xy+12=2012+12\Rightarrow (x+4)(y+3)=2024$$
by AM-GM inequality, we have
$$x+4+y+3=x+4+\frac{2024}{x+4}\ge 2\sqrt{2024}\Rightarrow x+y\ge 2\sqrt{2024}-7$$
Here, note that 
$$2024\lt 2025=45^2\Rightarrow \sqrt{2024}\lt 45\Rightarrow 2\sqrt{2024}-7\lt 83$$and that 
$$2024\gt 1980.25=(44.5)^2\Rightarrow \sqrt{2024}\gt 44.5\Rightarrow 2\sqrt{2024}-7\gt 82$$
Thus, we can say $$x+y\ge 83.$$
A: $$3x+4y+xy=2012$$
is equivalent to:
$$ (x+4)(y+3) = 2024 = 2^3\cdot 11\cdot 23$$
(that allows to find all the integer solutions) hence by the AM-GM inequality:
$$ \frac{(x+4)+(y+3)}{2}\geq \sqrt{(x+4)(y+3)} = \sqrt{2024}  $$
so:
$$ x+y+7 \geq 2\sqrt{2024} = 89.977775\ldots $$
and since $x,y$ are integers, that implies $x+y\geq 83$ as wanted. That minimim is attained by $x=42,y=41$ or $x=40,y=43$.
A: 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}
