Olympiad Math Question - If the sum of the positive inverses of 4 positive integers equals $1.1$, what’s the lowest possible sum of the integers? I was going through some Olympiad pass papers and came across this question:

Given four different positive integers $A, B, C, D$ so that $\frac{1}{A}+\frac{1}{B}+ \frac{1}{C}+\frac{1}{D}=1.1$. Find the smallest possible value of $A+B+C+D$.

Does the value $1.1$ have to do with anything? Also, what trick can I use to solve this question? Is there an inverse equation formula of some type I can use?
I tried doing this:
$$\frac{1}{A}+\frac{1}{B}+ \frac{1}{C}+\frac{1}{D}=1.1$$
$$ABC+BCD+ACD+ABD=1.1ABCD$$
But now I don’t know where to continue. Also, this is an Olympiad math question, which means I probably need an answer that can solve the question in 2 minutes or less.
 A: Without loss of generality, suppose $A < B < C < D$.
Notice that if $A \ge 3$, then $B \ge 4$, $C \ge 5$, $D \ge 6$, and then $\tfrac{1}{A}+\tfrac{1}{B}+\tfrac{1}{C}+\tfrac{1}{D} \le \tfrac{1}{3}+\tfrac{1}{4}+\tfrac{1}{5}+\tfrac{1}{6} = \tfrac{19}{20} < 1.1$. So we need $A = 1$ or $A = 2$.
Case 1: $A = 2$. Then we need $\tfrac{1}{B}+\tfrac{1}{C}+\tfrac{1}{D} = \tfrac{3}{5}$ with $2 < B < C < D$.
Since $\tfrac{3}{B} > \tfrac{1}{B}+\tfrac{1}{C}+\tfrac{1}{D} = \tfrac{3}{5}$, we must have $B < 5$, i.e. $B = 3$ or $B = 4$.
If $B = 4$, we need $\tfrac{1}{C}+\tfrac{1}{D} = \tfrac{7}{20}$ with $4 < C < D$. Since $\tfrac{2}{C} > \tfrac{1}{C}+\tfrac{1}{D} = \tfrac{7}{20}$, we must have $C < \tfrac{40}{7}$, i.e. $C \le 5$. Since $4 < C \le 5$, we must have $C = 5$, but then $D = \tfrac{20}{3}$, which is not an integer. So there are no solutions with $A = 2$ and $B = 4$.
If $B = 3$, we need $\tfrac{1}{C}+\tfrac{1}{D} = \tfrac{4}{15}$ with $3 < C < D$. Since $\tfrac{2}{C} > \tfrac{1}{C}+\tfrac{1}{D} = \tfrac{4}{15}$, we must have $C < \tfrac{15}{2}$, i.e. $C \le 7$. Testing $C = 4, 5, 6, 7$ yields $D = 60, 15, 10, \tfrac{105}{13}$ respectively. In this case, the smallest sum where $C$ and $D$ are integers is $21$ which occurs for $(A,B,C,D) = (2,3,6,10)$.
Case 2: $A = 1$. Then, we need $\tfrac{1}{B}+\tfrac{1}{C}+\tfrac{1}{D} = \tfrac{1}{10}$. But since $\tfrac{3}{D} < \tfrac{1}{B}+\tfrac{1}{C}+\tfrac{1}{D} = \tfrac{1}{10}$, any solution in this case will have $D > 30$, and thus, $A+B+C+D > 30 > 21$. So we will not find a smaller sum in this case.
Therefore, the minimum sum is $21$.

Note that if you just need to get an answer quickly without a rigorous proof, then you can probably just guess and check until you find something reasonably small. In problems with Egyptian fractions (fractions with numerator $1$), the sum $\tfrac{1}{2}+\tfrac{1}{3}+\tfrac{1}{6} = 1$ comes up a lot, namely it is the smallest set of distinct Egyptian fractions that add up to $1$. So it's not too hard to build off of that to get $\tfrac{1}{2}+\tfrac{1}{3}+\tfrac{1}{6}+\tfrac{1}{10} = \tfrac{11}{10}$. I'm not sure if there is an easy way to convince yourself that's the smallest sum though.
A: As your equivalent equation shows, at least one of the numbers is even, and at least one must be divisible by $5$. Since we are trying to make the numbers as small as possible, let's see if $2$ and $5$ will do the trick. Now $0.5+0.2=0.7$, leaving $0.4$ to make up with two Egyptian fractions (EFs). Starting with the smallest denominator, namely $3$, we are already lucky: $\frac13+\frac1{15}=0.4$. Can we do better than this? Our target to beat is $2+3+5+15=25$. So we are looking for four positive integers, including at least one even number and one multiple of $5$, which add to $24$ or less, whose reciprocals sum to $1.1$.
Clearly $1$ can be excluded, because that would leave us having to find three EFs summing to $0.1$, and the denominator sum would exceed $30$. Next comes $2$ (as before). That leaves a three-term EF sum of 0.6. It’s easy to spot $\frac13+\frac16+\frac1{10}=0.6$. Our integer sum target is now $20$ or less. Note that if any number has a prime-power factor not dividing $10$, then another has that factor too. Hence we can forget $7,8,9$, any of $11,12,13,14,15,16,17,18$, and $19$, and larger numbers, since the inclusion of any one would drive our sum over $20$. What is left to play with is $2,3,4,5,6$ and $10$. It is clear that
$$\frac12+\frac13+\frac16+\frac1{10}=1.1$$
cannot be bettered.
