If a, b, c, d and p are different real numbers such that : (a 2 + b 2 + c 2 )p 2 – 2(ab + bc + cd) p + (b 2 + c 2 + d 2 ) ≤ 0, then show that a, b, c and d are in G.P.

Question

Philoid · Accepted Answer

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 .

If a, b, c, d and p are different real numbers such that :(a2 + b2 + c2)p2 – 2(ab + bc + cd) p + (b2 + c2 + d2) ≤ 0, then show that a, b, c and d are in G.P.

If a, b, c, d and p are different real numbers such that :
(a² + b² + c²)p² – 2(ab + bc + cd) p + (b² + c² + d²) ≤ 0, then show that a, b, c and d are in G.P.