We have $$[m,n] + [m+1,n+1] = \frac{mn}{(m,n)} + \frac{(m+1)(n+1)}{(m+1,n+1)}.$$ Let $d_1 = (m,n)$ and $d_2=(m+1,n+1)$. It is clear that (a) $d_1,d_2\mid n-m$, and (b) $d_1$ and $d_2$ are coprime. So, $d_1 d_2 \mid n-m$ and therefore $d_1d_2\le n-m$. Using this fact, we get via AM-GM that $$\frac{mn}{d_1}+\frac{(m+1)(n+1)}{d_2}> mn\left(\frac{1}{d_1}+\frac{1}{d_2}\right)\ge \frac{2mn}{\sqrt{d_1d_2}} \ge \frac{2mn}{\sqrt{n-m}},$$ as claimed. Note that the inequality is strict.

Other solution: n=m+k gcd(m,m+k)*gcd(m+1,m+k+1)=gcd(k,m)*gcd(k,m+1)<=k (gcd(k,m)|k, gcd(k,m+1)|k, because gcd{gcd(k,m),gcd(k,m+1)}=1, gcd(k,m)*gcd(k,m+1)|k -> gcd(k,m)*gcd(k,m+1)<=k) lcm(m,n)=mn/gcd(m,n), lcm(m+1,n+1)=(m+1,n+1)/gcd(m+1,n+1) lcm(m,n)+lcm(m+1,n+1)>mn/gcd(m,n)+mn/gcd(m+1,n+1)>=2mn(1/gcd(m,n)*gcd(m+1,n+1))^(1/2)>=2mn/(k)^(1/2)=2mn/(m-n)^(1/2) So, lcm(m,n)+lcm(m+1,n+1)>=2mn/(m-n)^(1/2)

Let a>b>c be postive integers then prove that: lcm(a,b,c)+lcm(a+1,b+1,c+1)>=8abc/(a-c)^(3/2)