Prove for $m,\ n$ $\in \Bbb N$ and $m > n$ we have : $\text{lcm}(m,n)+\text{lcm}(m+1,n+1)>\frac{2mn}{\sqrt{m-n}}$

Question

I'll show lcm $(a,b)$ by $[a,b]$ and $\gcd(a,b)$ by $(a,b) $.

$$[m,n]+[m+1,n+1]>\frac{2mn}{\sqrt{n-m}} $$

This is my attempt:

\begin{align*} m-n&\ge (m,n)\\ \implies [m,n]&\ge\frac{mn}{m-n} \end{align*}

so we have :

\begin{align*} [m,n]+[m+1,n+1]&\ge\frac{mn+(m+1)(n+1)}{m-n}\\ &=\frac{2mn}{m-n}+\frac{m+n+1}{m-n} \end{align*}

and here I don't know how to continue.