Note: $(x,y) := \gcd(x, y)$
Problem:
Prove that $$(a+b, a-b) \ge (a,b)$$ for $a,b \in \mathbb{Z}$
Method:
I used this method which failed.
$$(a+b, a-b) = \min \{(a+b)x+(a - b)y | x,y \in \mathbb{Z}\} $$ $$ = \min \{a(x+y) + b(x-y)| x,y \in \mathbb{Z}\}$$ $$ = \min \{am + bn| m,n \in \mathbb{Z}\} = (a, b)$$
which is obviously not what I was supposed to prove :(
Please note that I was able to prove this question later using another method $\Rightarrow$
Let $(a,b) = d$ then $a = dk$ and $b = dl$ for some $k,l \in \mathbb{Z}$.
Now, $(a+b, a-b) = (d(k+l), d(k-l)) = d(k+l, k-l) \ge d = (a,b)$ Q.E.D.
Question:
What was the problem with my first method?
The two sets $$ A = \mathbb N \cap \left\{\, a(x+y) + b (x-y) \,\middle|\, x,y\in\mathbb Z\,\right\} $$ and $$ B = \mathbb N \cap \left\{\, an + bm \,\middle|\, n,m\in\mathbb Z\,\right\} $$ are not equal in general. Obviously $A\subseteq B$, since we can choose $n=x+y$ and $m=x-y$, which are both integers. However, in general $B$ is not a subset of $A$. Take $an+bm\in B$. If you try to have $n=x+y$ and $m=x-y$ for $x,y\in\mathbb Z$, you end up with $$ x = \frac{n+m}{2}, \quad y = \frac{n-m}{2}. $$ These $x$ and $y$ are not integers, if $n-m$ is not even!
From $A\subseteq B$ however, you can indeed conclude $\min A \ge \min B$, which is what you wanted to show.