If a, b, c, d and

We observe that the left side of the inequality could be written like:

(ap - b)² + (bp - c)² + (cp - d)² ≥ 0 as each of these 3 terms is a

perfect square..........(1)

But by the given condition:

(ap - b)² + (bp - c)² + (cp - d)² ≤ 0 ........(2)

Therefore the conditions (1) and(2) can be satisfying iff the sum, (ap - b)² + (bp - c)² + (cp - d)² = 0 which is possible iff each of the terms,

(ap - b) = (bp - c) = (cp - d) is equal to zero. So,

ap - b = 0, Or a/b = k.

bp - c = 0. Or b/c = k.

cp - d = o, Or a/d =k.

Therefore, a/ b = b/c = c/d = k the common ratio of the terms a, b, c, and d.

So a,b,c,and d are in geometric progression .