If `a,b,c,d` be four distinct positive quantities in GP,then show that
(a) `a+dgtb+c`

Explanation

`therefore a,b,c,d` are in GP.
(a) Applying AMgtGM
For first three members,
`(a+c)/(2)gtb`
`implies a+cgt2b " " "….(v)"`
For last three members, `(b+d)/(2)gtc`
`implies b+dgt2c " " "…..(vi)"`
From Eqs. (v) and (vi), we get
`a+c+b+dgt2b+2c " or " a+dgtb+c`
(b) Applying GMgt HM
For first three members, `bgt(2ac)/(a+c)`
`implies ab+bcgt 2ac " " "....(vii)"`
For last three members, `cgt(2bd)/(b+d)`
`implies bc+cdgt2bd " " "....(viii)"`
From Eqs. (vii) and (viii), we get
`ab+bc+bc+cdgt2ac+2bd`
or `ab+cdgt 2(ac+bd-bc)`
Dividing in each term by `abcd`, we get
` c^(-1)d^(-1)+a^(-1)b^(-1)gt2(b^(-1)d^(-1)+a^(-1)c^(-1)-a^(-1)d^(-1))`