Prove $\gcd(ab, c)\le \gcd(a,c)\gcd(b,c)$ How do we prove that 
$$\gcd(ab, c) \le \gcd(a,c) \gcd(b,c)$$
for all positive integers $a,b,c$? I've tried thinking about it with Bezout's and the Euclidean algorithm but I'm not quite getting there. Very new to proof-based math so any help is very appreciated.
 A: HINT:
Write 
$$\operatorname{gcd}(a,c) = s a + t c \\
\operatorname{gcd}(b,c)= p b + q c $$ and multiply the equalities, then see that the RHS is divisible by $\operatorname{gcd}(ab,c)$.
A: When in doubt with $\gcd$ problems, use prime factorization. While it may not be the most abstract or "elegant" way to prove things, it is very powerful and will get the job done.
So, we can write
\begin{align*}
a &= 2^{x_1} 3^{x_2} 5^{x_3} \cdots \\
b &= 2^{y_1} 3^{y_2} 5^{y_3} \cdots \\
c &= 2^{z_1} 3^{z_2} 5^{z_3} \cdots \\
ab &= 2^{x_1 + y_1} 3^{x_2 + y_2} 5^{x_3 + y_3} \cdots 
\end{align*}
Now, when we take the gcd, we take the minimum of the two prime exponents, for each prime. So
$$
\gcd(ab,c) = 2^{\min(x_1 + y_1,z_1)} 3^{\min(x_2 + y_2, z_2)} \cdots,
$$
and in general the power of a prime $p$ here is $\min(x_i + y_i, z_i)$.
On the other hand,
\begin{align*}
\gcd(a,c) &= 2^{\min(x_1,z_1)} 3^{\min(x_2,z_2)} \cdots \\
\gcd(b,c) &= 2^{\min(y_1,z_1)} 3^{\min(y_2,z_2)} \cdots \\
\gcd(a,c) \gcd(b,c) &=
\left( 2^{\min(x_1,z_1)} 3^{\min(x_2,z_2)} \cdots \right)
\left( 2^{\min(y_1,z_1)} 3^{\min(y_2,z_2)} \cdots \right) \\
&= 2^{\min(x_1,z_1) + \min(y_1,z_1)} 3^{\min(x_2,z_2) + \min(y_2,z_2)} \cdots,
\end{align*}
and in general the power of a prime $p$ here is $\min(x_i, z_i) + \min(y_i,z_i)$.
So how can we prove the desired result?
We can do it by proving that the prime powers of $\gcd(ab,c)$ are less than the prime powers of $\gcd(a,c) \gcd(b,c)$.
That is, we just have to show
$$
\min(x_i + y_i, z_i) \le \min(x_i,z_i) + \min(y_i,z_i).
$$
Now this can be shown true by considering whether $x_i + y_i$ or $z_i$ is bigger.
