Greatest Common Divisor of a+b,a-b

Question

Prove that $gcd(a + b, a − b) ≥ gcd(a, b)$

Let $d=gcd(a+b,a-b)$
So $d=m(a+b)+n(a-b) = a(m+n)+b(m-n)$
Which implies $d|a$ and $d|b$
Therefore, $d|gcd(a,b)$
$gcd(a,b)=dx ≥ d= gcd(a + b, a − b)$

Why am I getting the opposite inequality?

Answer 1

$$\gcd(a,b)\mid a,b\implies \gcd(a,b)\mid a+b,a-b$$

$$\iff \gcd(a,b)\mid \gcd(a+b,a-b)\implies \gcd(a,b)\le \gcd(a+b,a-b)$$

Answer 2

Let $g=\gcd(a,b)$. Then $$ \begin{align} \gcd(a+b,a-b) &=g\gcd\left(\frac{a+b}g,\frac{a-c}g\right)\\ &\ge g \end{align} $$ since $$ \gcd\left(\frac{a+b}g,\frac{a-c}g\right)\ge1 $$