May I know if my solution is correct? Is there a more elegant way to tackle this problem? Thank you very much.
By Cauchy schwarz inequality, $(a+b+c)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right) \ge9 \Rightarrow a+b+c\geq 378,$ with equality if $a=b=c=126.$
Let $c=126,$ then $\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{63}.$ By Cauchy schwarz, $(a+b)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\geq 4 \Rightarrow a+b\geq 252.$
Assume $ b < 252.$ Since $a=\dfrac{63^2}{b-63}+63$ it follows that $ b-63 < 189$ and $(b-63) \mid 63^2.$ Checking the factors of $63^2$ no bigger than $189$, $\text{min}(a+b) = 144+112=256.$ We can check if there exists distinct $a, b$ such that $a+b = 253, 254$ or $255,$ then $\text{min}(a+b)=256$ again.
Next, let us show that $\text{min}(a+b+c) = 144+112+126=382.$ It suffices to show there exists no integer solution to $a+b+c = 381, 380$ or $379.$
Let $a+b+c = S$. Assume wlog $43\leq a < 126.$ $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{42} \Leftrightarrow b^2-(S-a)b+\dfrac{42a(S-a)}{a-42}=0.$
We require the determinant $D=(S-a)^2+\dfrac{168a(S-a)}{a-42} \geq 0.$
For $S=381,$ $D\geq 0 \Leftrightarrow \dfrac{255}{2} - \dfrac{\sqrt{1017}}{2}< 112\leq a \leq 142 < \dfrac{255}{2} + \dfrac{\sqrt{1017}}{2}.$ Since $a<126,$ it follows that $112 \leq a \leq 125.$
We can rewrite $D=(381-a)^2+168a-339\times 168-\dfrac{339\times 168\times 42}{a-42}.$
For $a\in \{112, ...,125\} \setminus \{114\}, (a-42) \in \{70, ...,83\} \nmid 339 \times 168\times 42,$ hence $b \not\in \mathbb{Z}.$
If $a = 114, \ D = 267 \Rightarrow \sqrt{D} \not\in \mathbb{Q} \Rightarrow b \not \in \mathbb{Z}.$
For $S = 380, $ following previous argument, we require $D = (380-a)^2+168a-338\times 168-\dfrac{338\times 168 \times 42}{a-42} \geq 0, $ so $114\leq a \leq 125.$
For $a\in \{115, ...,125\} \setminus \{114\}, (a-42) \nmid 338 \times 168\times 42,$ hence $b \not\in \mathbb{Z}.$
If $a=114, \ b = c = 133.$ But $b,c$ are distinct.
We can argue similarly that there are no integer solutions to $a+b+c = 379.$
