a) We will show the inequality is true by showing that the inequality can't be false. Suppose it is false. We have $$\frac{1}{[a, b]} + \frac{1}{[b, c]} + \frac{1}{[c, d]} + \frac{2}{[d, e]} > 1$$ Now, notice that if $x<y$ positive integers, then $[x,y] > x$ clearly. Therefore, we establish $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{2}{d} > \frac{1}{[a, b]} + \frac{1}{[b, c]} + \frac{1}{[c, d]} + \frac{2}{[d, e]} > 1$$Now, as $a = min(a,b,c,d)$, we even have $$\frac{5}{a} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{2}{d} > \frac{1}{[a, b]} + \frac{1}{[b, c]} + \frac{1}{[c, d]} + \frac{2}{[d, e]} > 1$$This is clearly false if $a \ge 5$ so $a \le 4$. Notice that if $a=4$, then the inequality is still false because $$1 > \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{2}{7} \ge > \frac{1}{a} + \frac{1}{a+1} + \frac{1}{a+2} + \frac{2}{a+3} \ge \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{2}{d} > 1$$This implies that $a=1,2,3$. Now, we have to bash these cases, but they are not so hard. I will leave that up to the reader.

Alternatively, we may note that $[x,y]=\dfrac{xy}{(x,y)} \leq \dfrac{xy}{x-y}$ and $[x,y] \geq 2y$ for all $x>y$, hence $\dfrac{1}{[a,b]}+\dfrac{1}{[b,c]}+\dfrac{1}{[c,d]}+\dfrac{2}{[d,e]} \leq (\dfrac{1}{a}-\dfrac{1}{b})+(\dfrac{1}{b}-\dfrac{1}{c})+(\dfrac{1}{c}-\dfrac{1}{d})+\dfrac{2}{2d}=\dfrac{1}{a} \leq 1,$ as desired. Eqality holds if $a=1$ and $(a,b)=b-a, (b,c)=c-b,(c,d)=d-c$ and $[d,e]=2d$. The last equality can only hold for $e=2d.$ Moreover, $b-1=(b,a)=(a,b)=(1,b)=1$, i.e. $b=2$. Now, if $c$ is odd then $c=3$, while if $c$ is even then $c=4$. If $c=3$ then $(3,d)=d-3$ and so $d \in \{4,6 \},$ while if $c=4$ then $(4,d)=d-4$ and so $d \in \{5,6, 8 \}$. Hence, there is a total of $5$ triples attaining equality: $(a,b,c,d,e)=(1,2,3,4,8),(1,2,3,6,12),(1,2,4,5,10),(1,2,4,6,12),(1,2,4,8,16)$

another intresting idea we can do casework like that : if $lcm(a,b) \geq 5$ the inequality is true Proof: $$\dfrac{1}{[a,b]}+\dfrac{1}{[b,c]}+\dfrac{1}{[c,d]}+\dfrac{2}{[d,e]}$$$\dfrac{1}{[a,b]} \le \frac{1}{5}$ $\dfrac{1}{[c,b]} \le \frac{1}{5}$ $\dfrac{1}{[c,d]} \le \frac{1}{5}$ $\dfrac{2}{[d,e]} \le \frac{2}{5}$ If each of them are true so inequality is true but we have $a < b < c < d < e$ condition and if we prove only $lcm(a,b) \geq 5$ we are done now we have a few cases such as $lcm(a,b)={2,3,4}$ besides $lcm(b,c)=4$ and that is easy to find out

that's my solution's motivation during exam from a silly mistake i didn't participate at JBMO (1 point remain to the goal ...... )

Claim: $\frac{1}{a}-\frac{1}{b} \ge \frac{1}{[a,b]}.$ Multiplying, both sides, by $ab,$ gives, $b-a \ge \frac{ab}{[a,b]}=\gcd(a,b).$ It, is quite obvious that $b-a \ge \gcd(a,b).$ Now, we have that $\frac{1}{[a, b]} + \frac{1}{[b, c]} + \frac{1}{[c, d]} + \frac{2}{[d, e]}=\frac{1}{a},$ by telescoping, and clearly, $\frac{1}{a} \le 1,$ equality, holds, when $a=1.$

Lukaluce wrote: Let $a < b < c < d < e$ be positive integers. Prove that $$\frac{1}{[a, b]} + \frac{1}{[b, c]} + \frac{1}{[c, d]} + \frac{2}{[d, e]} \le 1$$where $[x, y]$ is the least common multiple of $x$ and $y$ (e.g., $[6, 10] = 30$). When does equality hold? VERY GREAT PROBLEM LOVED IT

$\boxed{ANSWER:a=1}$ claim:$\frac{1}{a}-\frac{1}{b}\geq \frac{1}{lcm(a,b)}$ proof:$b-a\geq gcd(a,b)=gcd(b-a,a)$ $1\geq\frac{1}{a}-\frac{1}{b}+\frac{1}{b}-\frac{1}{c}+\frac{1}{c}-\frac{1}{d}+\frac{1}{d}=\frac{1}{a}\geq LHS$ since $a\geq 1\implies 1\geq \frac{1}{a}$