Easiest and most complex proof of $\gcd (a,b) \times \operatorname{lcm} (a,b) =ab.$

Let $\gcd(a,b)=d$. Then for some $a_0,b_0$ such that $a_0$ and $b_0$ are relatively prime, we have $a=da_0$ and $b=d b_0$. If we can show that the lcm of $a$ and $b$ is $da_0b_0$, we will be finished.

Certainly $da_0b_0$ is a common multiple of $a$ and $b$. We must show that it is the least common multiple.

Let $m$ be a common multiple of $a$ and $b$. We will show that $da_0b_0$ divides $m$.

Since $m$ is a multiple of $a$, we have $m=ka=ka_0d$ for some $k$. But $b$ divides $m$, so $db_0$ divides $ka_0d$, and therefore $b_0$ divides $ka_0$. Since $a_0$ and $b_0$ are relatively prime, it follows that $b_0$ divides $k$, and we are finished.

First notice that $$ \dfrac{ab}{\gcd(a,b)} = a\dfrac{b}{\gcd(a,b)} = b\dfrac{a}{\gcd(a,b)} $$ is a common multiple of $a$ and $b$. By the minimality of the $\operatorname{lcm}$, $$ \frac{ab}{\gcd(a,b)}\ge\operatorname{lcm}(a,b)\Longrightarrow ab\ge\operatorname{lcm}(a,b)\gcd(a,b)\tag{1} $$ By division, we can write $$ ab = q\operatorname{lcm}(a,b) + r\quad\text{where}\quad0 \le r \lt \operatorname{lcm}(a,b) $$ Because $ab$ and $\operatorname{lcm}(a,b)$ are common multiples of $a$ and $b$, so is $r$. By the minimality of the $\operatorname{lcm}$, $r = 0$. Therefore, $\operatorname{lcm}(a,b)$ divides $ab$. Notice that $$ \frac{ab}{\operatorname{lcm}(a,b)} = \frac{a}{\operatorname{lcm}(a,b)/b} = \frac{b}{\operatorname{lcm}(a,b)/a} $$ is a common divisor of $a$ and $b$. By the maximality of the $\gcd$, $$ \frac{ab}{\operatorname{lcm}(a,b)} \le \gcd(a,b)\Longrightarrow ab\le\operatorname{lcm}(a,b)\gcd(a,b)\tag{2} $$ Combining $(1)$ and $(2)$, we get $$ ab = \operatorname{lcm}(a,b)\gcd(a,b) $$