I came across this question while solving the book Challenge and Thrill of Pre-College mathematics. Please check if the proof that I have done is correct. (I am familiar with only basic number theory including basic theory of primes and modular arithmetic.)
I believe that I might have made some mistake in reasoning while proving that $bx-ay\geq 0$
Proof:
Without loss of generality, we can assume that $a$ and $b$ are both positive with $a\geq b \gt 0$
Since GCD can be written as linear combination of the two integers,
$$(a,b) = ax + by$$
We know that $(a,b) \leq b \leq a$ Which implies that either $x$ or $y$ is is negative.
We can assume that $y$ is negative. This implies $bx-ay\geq 0$. (Conversely if $x$ is negative then $ay-bx \geq 0$ )
Now,
$$ax + by = (a,b)$$ $$\Rightarrow ax + by + bx - ay \geq (a,b)$$ $$\Rightarrow (a+b)x + (b-a)y \geq (a,b)$$ $$\Rightarrow (a+b)x + (-y)(a-b) \geq (a,b)$$ $$\Rightarrow (a+b, a- b) \geq (a,b)$$
QED.
Let $d=\gcd(a+b,a-b)$ and $\gcd(a,b)=e$.
$e$ must divide the sum and the difference of $a$ and $b$. This means that $e \mid a+b$ and $e \mid a-b$. Thus $e$ is a common divisor of $a+b$ and $a-b$. Thus $e \mid \gcd(a+b, a-b)=d$. Thus
$$e \leq d.$$
Your question is answered.
However one can do a bit more:
Furthermore, $d \mid a+b$ and $d \mid a-b$. Then $d \mid 2a$ and $d \mid 2b$. Thus $d$ is a common divisor of $2a$ and $2b$. Consequently, $d \mid \gcd(2a,2b)$. But $\gcd(2a,2b)=2\gcd(a,b)=2e$. Thus, $$d \mid 2e \implies d \leq 2e.$$
So we can say $$ e \leq d \leq 2e.$$