Given that $a,b,c,d$ and $e$ are positive integers such that $a<b<c<d<e$. Find the maximum value of $$\frac1{[a,b]} +\frac1{[b,c]} +\frac1{[c,d]} + \frac1{[d,e]},$$ where $[x,y]$ denotes the LCM of $x$ and $y$
Taking $b = 2a$ Hence, $$\frac{1}{\text{lcm}(1,2)}+\frac{1}{\text{lcm}(2,4)}+\frac{1}{\text{lcm}(4,8)}+\frac{1}{\text{lcm}(8,16)}=\frac{15}{16}$$ So can anyone come up with valid proof that minimum value of $[a,b]$ = $[a,2a]$
If you want to minimize $[a,b]$ therefore, you can simply choose $b = 2a$ ,Note : b = a would have been the minimum but since $b>a$ and $b \neq a$ but so we have to look for the next integer multiple of which 2 is so $b = 2a$, since $b>a$ and $b \neq a$
Hence, $$\frac{1}{\text{lcm}(1,2)}+\frac{1}{\text{lcm}(2,4)}+\frac{1}{\text{lcm}(4,8)}+\frac{1}{\text{lcm}(8,16)}=\frac{15}{16}$$
Thanks to @астонвіллаолофмэллбэрг and Raffaele