1u mean to say this
(a2x2 -2abx+b2 ) + (b2x2 - 2cbx +c 2) +(c2x2 - 2cdx + d2)
(ax-b)2 + (bx-c)2+ (cx-d)2 ≤0
now wat to do
if a,b,c,d>0, x e R and (a2 + b2 +c 2)x2 -2(ab+bc+cd)x + b2+ c2 +d2 ≤0
then the value of the determinant
[33 14 lna]
[65 27 lnb]
[97 40 lnc]
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4 Answers
62Hint: Prove that a, b, c, and d are in GP
That can be done by expressing the original sum as the sum of 3 perfect squares.
[1]
Can you solve it now?
162now because they are all real and positive
for the inequality to hold, each will have to be zero..
so ax=b, bx=c and cx=d
hence you will also get that lna, ln b, ln c and ln d are in AP
Now proceed to compute the matrix.. (*I havent myself.. but will be surprised if the answer is not zero)