Question

If a, b, c and d are in proportion, prove that: `abcd [(1/a^2 + 1/b^2 + 1/c^2 + 1/d^2]` = a² + b² + c² + d² 

Answer 1

∵ a, b, c, d are in proportion
`a/b = c/d` = k(say)
a = bk, c = dk.
L.H.S. = `abcd (1/a^2 + 1/b^2 + 1/c^2 + 1/d^2)`

= `bk.b.dk.d [1/(b^2k^2) + 1/b^2 + 1/(d^2k^2) + 1/d^2]`

= `k^2b^2d^2 [(d^2 + d^2k^2 + b^2 + b^2k^2)/(b^2d^2k^2)]`

= d²(1 + k²) + b²(1 + k2)
= (1 + k²)(b² + d²)
R.H.S. = a² + b² + c2 + d²
= b²k² + b² + d²k² + d²
= b²(k² + 1) + d²(k² + 1)
= (k² + 1)(b² + d²)
∴ L.H.S. = R.H.S.