If $a, b, c$ and $d$ are in $G.P.$ Therefore,

$b c=a d$         ……….$(1)$

$b^{2}=a c$          ………$(2)$

$c^{2}=b d$         ………$(3)$

It has to be proved that,

$\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}$

$R.H.S.$

$=(a b+b c+c d)^{2}$

$=(a b+a d+c d)^{2}$           [ Using $(1)$ ]

$=[a b+d(a+c)]^{2}$

$=a^{2} b^{2}+2 a b d(a+c)+d^{2}(a+c)^{2}$

$=a^{2} b^{2}+2 a^{2} b d+2 a c b d+d^{2}\left(a^{2}+2 a c+c^{2}\right)$       [ Using $(1)$ and $(2)$ ]

$=a^{2} b^{2}+2 a^{2} c^{2}+2 b^{2} c^{2}+d^{2} a^{2}+2 d^{2} b^{2}+d^{2} c^{2}$

$=a^{2} b^{2}+a^{2} c^{2}+a^{2} c^{2}+b^{2} c^{2}+b^{2} c^{2}+d^{2} a^{2}+d^{2} b^{2}+d^{2} b^{2}+d^{2} c^{2}$

$=a^{2} b^{2}+a^{2} c^{2}+a^{2} d^{2}+b^{2} \times b^{2}+b^{2} c^{2}+b^{2} d^{2}+c^{2} b^{2}+c^{2} \times c^{2}+c^{2} d^{2}$

[ Using $(2)$ and $(3)$ and rearranging terms ]

$=a^{2}\left(b^{2}+c^{2}+d^{2}\right)+b^{2}\left(b^{2}+c^{2}+d^{2}\right)+c^{2}\left(b^{2}+c^{2}+d^{2}\right)$

$=\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=$ $L.H.S$

$\therefore L .H.S. = R . H.S.$

$\therefore\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}$