I'll use AM-GM with $lcm(n, m)$ and $lcm(n+1, m+1)$
$lcm(n, m) + lcm(n+1, m+1) \geq 2\sqrt{lcm(n, m) \cdot lcm(n+1, m+1)}$
Now we have to prove $2\sqrt{lcm(n, m) \cdot lcm(n+1, m+1)} \geq 2m\sqrt{n}$
$lcm(n, m) \cdot lcm(n+1, m+1) = \frac{nm}{gcd(n, m)} \cdot \frac{(n+1)(m+1)}{gcd(n+1, m+1)}$
Using euclidian algorithm $\frac{nm}{gcd(m, n-m)} \cdot \frac{(n+1)(m+1)}{gcd(m+1, n-m)}$
Since $m$ and $m+1$ are coprime, this equals $\frac{nm(n+1)(m+1)}{gcd(m(m+1), n-m)}$ also $n-m \geq gcd(m(m+1), n-m)$
Substuting, we have to prove that $\sqrt{\frac{nm(n+1)(m+1)}{gcd(m(m+1), n-m)}} \geq m\sqrt{n}$
Squaring, and using the fact from above: $\frac{nm(n+1)(m+1)}{n-m} \geq m^2n$
Since $a+1 \geq a$ we can substitute $\frac{m^2n^2}{n-m} \geq m^2n$
$\frac{n}{n-m} \geq 1$
true because $n > m$