GCD domain is LCM domain On this Wiki page it is written: 

A GCD domain is an integral domain $R$ with the property that any two non-zero elements have a greatest common divisor (GCD). Equivalently, any two non-zero elements of $R$ have a least common multiple (LCM).

How to prove last statement that is equivalence of GCD and LCM for all elements? (I am able to prove in Bezout ring but I am not able to prove in general GCD ring.)
Does the existence of gcd of two elements implies existence of lcm and conversely in any integral domain?

I want to make a few supplements to the above good answer by Xam.

If $R$ is an integral domain and $\gcd(a,b)$ exists, then $\mathrm{lcm}(a,b)$ may not exist.

Consider $\mathbb Z[\sqrt{-3}]$. Let $a = 2$ and $b = 1+\sqrt{-3}$.
If $\mathrm{lcm}(2,1+\sqrt{-3})$ exists, then $\mathrm{lcm}(2,1+\sqrt{-3})\mid 2(1+\sqrt{-3})$ and $\mathrm{lcm}(2,1+\sqrt{-3}) \mid 2(1-\sqrt{-3})$ (because $2(1-\sqrt{-3}) = -(1+\sqrt{-3})(1+\sqrt{-3})$). However, the factors of $2(1+\sqrt{-3})$ are $2, (1+\sqrt{-3})$ and $(1-\sqrt{-3})$.
Thus $\mathrm{lcm}(2,1+\sqrt{-3})$ must be equal to $2(1+\sqrt{-3})$. In the same way, according to $\mathrm{lcm}(2,1+\sqrt{-3})\mid 2(1-\sqrt{-3})$, we have $\mathrm{lcm}(2,1+\sqrt{-3})$ must be equal to $2(1-\sqrt{-3})$. This is a contradiction.
Actually, $\gcd(4,2(1+\sqrt{-3}))$ doesn't exist，but $\gcd(2,1+\sqrt{-3}) =1$.
