a,b,c,d are all real numbers and if (a2+b2+c2)p2 -2(ab+bc+cd)p + b2+c2+d2 <= 0 then prove that a,b,c,d are in GP

hi rohit

the discriminant of the equation cant be -ve as the leading coeff is +ve so its an upward parabola

which cuts at 2 points in x axis

now finding the discriminant

(ab+bc+cd)²   - (a2+b2+c2 )  (b2+c2+d2)     > = 0

but the L.H.S is not greater than 0 because of CAUCHY SCHWARZ INEQUALITY

so it can only be 0

so ∑ a²  ∑ b²   = ∑ ab                   applying for a , b ,c  and b , c, d

and in equality is true only when the terms involved in the sigma sign are proportional

i.e   a/b = b/c = c/d     so proved they are in G.P

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