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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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