$$2\gcd(a, b) \le \gcd(2a,2b)$$ I'm pretty sure it has something to do with Bezout's lemma or it could use prime factorization, but unsure on how to go exactly about it. Please help!
Full Disclosure: This came up on a midterm and something like this may be on the final so I want to make sure I have it down for next time.
On
It is the special case $\,d = \gcd(a,b)\,$ below.
$$ d\mid a,b\,\Rightarrow\, 2d\mid 2a,2b\,\overset{\!\rm\color{#c00}U}\Rightarrow\, 2d\mid \gcd(2a,2b)\Rightarrow\ 2d\le \gcd(2a,2b)\qquad$$
where we used $\rm\color{#c00}U$ = gcd Universal Property below (provable by Bezout as in the link)
$$ x\mid y,z\!\!\overset{\rm\color{#c00}U\!\!}\iff x\mid \gcd(y,z)\qquad$$
Alternatively employ $\,2\gcd(a,b) = \gcd(2a,2b)\,$ by the gcd Distributive Law.
Normally you would prove the equality, as suggested by the link provided in comments. In this case however, you only need to prove one inequality. For this notice $2\gcd(a,b)|2a$ (because $2|2$ and $\gcd(a,b)|a$ by definition) and similarly $2\gcd(a,b)|2b$. So by definition $2\gcd(a,b)|\gcd(2a,2b)$, which implies $2\gcd(a,b)\leq \gcd(2a,2b)$.