If $$ a, b, c \in N $$ , and $$ 1\leq a\leq b\leq c $$ , find all $$ a,b,c $$ numbers that hold the equation always true,
$$ \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{3}{4} $$

Question

If $$ a, b, c \in N $$ , and $$ 1\leq  a\leq  b\leq  c $$ , find all $$ a,b,c $$ numbers that hold the equation always true,
$$ \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{3}{4}  $$

Uniteasy Admin · Accepted Answer

Case 1, if $$ a\geq 5 $$
$$ 5\leq a\leq b\leq c $$
$$ 	herefore \dfrac{1}{5}\geq \dfrac{1}{a} \geq \dfrac{1}{b} \geq \dfrac{1}{c} $$
$$ \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{ 1}{c}\leq \dfrac{3}{5}< \dfrac{3}{4}  $$
So no solution if $$ a\geq 5 $$ 
Case 2, if $$ a=1 $$
$$ \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}>\dfrac{3}{4} $$ , no solution.
Case 3, if $$ a=4 $$
$$ \dfrac{1}{b}+\dfrac{1}{c}=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{1}{2} $$
Case 3-1, if $$ b\geq 5 $$ 
$$ 5\leq b\leq c $$
$$ \dfrac{1}{5}\geq \dfrac{1}{b} \geq \dfrac{1}{c} $$
$$ 	herefore \dfrac{1}{b}+\dfrac{1}{c} <\dfrac{2}{5} <\dfrac{1}{2} $$ , no solution 
Case 3-2, if $$ b=1 $$
$$ \dfrac{1}{c}=-\dfrac{1}{4} $$ , no solution
Case 3-3, if $$ b=4 $$
$$ \dfrac{1}{c}=\dfrac{1}{2}-\dfrac{1}{4} $$
$$ 	herefore c=4 $$
Now we get our first solution $$ a=b=c=4 $$
Case 3-4, if $$ b=3 $$
$$ \dfrac{1}{c}=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6} $$
$$ 	herefore c=6 $$
So $$ a=4 $$ , $$ b=3 $$ $$ ,c=6 $$ is the 2nd solution we find.
Case 3-5, if $$ b=2 $$
$$ \dfrac{1}{c}=\dfrac{1}{2}-\dfrac{1}{2}=0 $$ , no solution 
Case 4, if $$ a=3 $$
$$ \dfrac{1}{b}+\dfrac{1}{c}=\dfrac{3}{4}-\dfrac{1}{3}=\dfrac{5}{12} $$ 
Case 4-1, if $$ b\geq 6 $$
$$ 6\leq b\leq c $$
$$ 	herefore \dfrac{1}{b}+\dfrac{1}{c} <\dfrac{2}{6} <\dfrac{5}{12} $$ , no solution
Case 4-2, if $$ b=1 $$ 
$$ \dfrac{1}{c} =\dfrac{5}{12} -1=-\dfrac{7}{12} $$ , no solution 
Case 4-3, if $$ b=5 $$
$$ \dfrac{1}{c} =\dfrac{5}{12}-\dfrac{1}{5} $$
$$ =\dfrac{25-12}{60} $$ , c is not an integer, no solution 
Case 4-4, if $$ b=4 $$
$$ \dfrac{1}{c} =\dfrac{5}{12}-\dfrac{1}{4} =\dfrac{1}{6} $$
$$ 	herefore c=6 $$
So $$ a=3 $$ , $$ b=4 $$ , $$ c=6 $$ is another solution
Case 4-5, if $$ b=3 $$
$$ \dfrac{1}{c} =\dfrac{5}{12} -\dfrac{1}{3} =\dfrac{1}{12} $$ 
$$ 	herefore c=12 $$
So $$ a=3 $$ , $$ b=3 $$ , $$ c=12 $$ is another solution 
Case 4-6, $$ b=2 $$
$$ \dfrac{1}{c} =\dfrac{5}{12}-\dfrac{1}{2}=-\dfrac{1}{12} $$ , no solution 
Case 5, $$ a=2 $$
$$ \dfrac{1}{b}+\dfrac{1}{c}=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4} $$
Case 5-1, if $$ b\geq 9 $$ 
$$ 9\leq b\leq c $$
$$ \dfrac{1}{9}\geq \dfrac{1}{b}\geq \dfrac{1}{c} $$
$$ \dfrac{1}{b}+\dfrac{1}{c} \leq \dfrac{2}{9}< \dfrac{1}{4} $$
$$ 	herefore  b\geq9 $$ , no solution
Case 5-2, $$ b\leq4 $$
$$ \dfrac{1}{c}=\dfrac{1}{4}-\dfrac{1}{b}\leq 0 $$ , no solution
Case 5-3, $$ b=8 $$
$$ \dfrac{1}{c}=\dfrac{1}{4}-\dfrac{1}{8} =\dfrac{1}{8} $$
$$ 	herefore c=8 $$
So $$ a=2 $$ , $$ b=8 $$ , $$ c=8 $$ is another solution 
Case 5-4, $$ b=7 $$
$$ \dfrac{1}{c}=\dfrac{1}{4}-\dfrac{1}{7} =\dfrac{3}{28} $$ 
$$ c $$ is not an integer, no solution 
Case 5-5, $$ b=6 $$
$$ \dfrac{1}{c}=\dfrac{1}{4}-\dfrac{1}{6} =\dfrac{1}{12} $$
$$ 	herefore c=12 $$
So $$ a=2 $$ , $$ b=6 $$ , $$ c=12 $$ is another solution 
Case 5-6, $$ b=5 $$
$$ \dfrac{1}{c}=\dfrac{1}{4}-\dfrac{1}{5} =\dfrac{1}{20} $$ 
$$ 	herefore c=20 $$
So $$ a=2 $$ , $$ b=5 $$ , $$ c=20 $$ is another solution

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