have 1/y+1/z=a/x,1/x+1/z=b1/y,1/y+1/x=c1/z .This leads to (a+1)/x=(b+1)/y=(c+1)/z=k, and then to 1/(a+1)+1/(b+1)+1/(c+1)=1.Since x,y,z need to be distinct distinct, so are a,b,c , hence the only solution is 2,3,6. the question was: Find three distinct positive integers with the least possible sum such that the sum of the reciprocals of any two integers among them is an integral multiple of the reciprocal of the third integer. Plse make me understand the solution fast.need it immediately.

							
Dear student,
The solutionis quite clear..
1/y+1/z=a/x,1/x+1/z=b1/y,1/y+1/x=c1/z  .This leads to (a+1)/x=(b+1)/y=(c+1)/z=k, and then to  1/(a+1)+1/(b+1)+1/(c+1)=1.Since x,y,z need to be distinct  distinct, so are a,b,c , hence the only solution is 2,3,6
 






















































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