Let `a ,b` be positive real numbers. If `a A_1, A_2, b` be are in arithmetic progression `a ,G_1, G_2, b` are in geometric progression, and `a ,H_1, H_2, b` are in harmonic progression, show that `(G_1G_2)/(H_1H_2)=(A_1+A_2)/(H_1+H_2)=((2a+b)(a+2b))/(9a b)`

Explanation

Clearly, `A_(1)+A_(2)=a+b`
`1/(H_(1))+1/(H_(2))=1/a+1/b`
`rArr(H_(1)+H_(2))/(H_(1)H_(2))=(a+b)/(ab)`
`rArr(H_(1)+H_(2))/(H_(1)H_(2))=(A_(1)+A_(2))/(G_(1)G_(2))`
`rArr(G_(1)G_(2))/(H_(1)H_(2))=(A_(1)+A_(2))/(H_(1)+H_(2))`
Also, `1/H_(1)=1/a+1/3(1/b-1/a)`
`rArrH_(1)=(3ab)/(2b+a)`
and `1/(H_(2))=1/a+2/3(1/b-1/a)`
`rArrH_(2)=(3ab)/(2a+b)`
`rArr(A_(1)+A_(2))/O(H_(1)+H_(2))=(a+b)/(3ab(1/(ab+a)+1/(2a+b))`
`((2b+a)(2a+b))/(9ab)`