Question

$\huge\red{\fbox{\bf{Question}}}$

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

Answer 1

Step-by-step explanation:

Given (a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) ≤ 0

(a2 p2 + b2 p2 + c2 p2 )- (2abp + 2bcp + cpd) + (b2 + c2 + d2) ≤ 0

(a2 p2 – 2abp + b2) + (b2p2 – 2bpc + c2) + (c2p2 – 2cdp + d2) ≤ 0

(ap -b)2 + (bp – c)2 + (cp – d)2 ≤ 0

Sum of squares cannot be negative.

So (ap -b) = 0

ap = b …(i)

(bp – c) = 0

bp = c …(ii)

cp – d = 0

cp = d …(iii)

From (i) and (ii)

b/a = c/b

=> b2 = ac

From (ii) and (iii)

c/b = d/c

c2 = bd

Hence a, b, c, d are in GP.