If `f(x)=(a_1x+b_1)^2+(a_2x+b_2)^2+...+(a_n x+b_n)^2` , then prove that `(a_1b_1+a_2b_2++a_n b_n)^2lt=(a1 2+a2 2++a n2)^(b1 2+b2 2++b n2)dot`

Explanation

Given,
`f(x) = (a_(1)x + b_(1))^(2) + (a_(2)x+b_(2))^(2) + ...+ (a_(n)x + b_(n))^(2)` (1)
or `f(x) = (a_1^(2) + a_(2)^(2) + ... + a_(n)^(2))x^(2) + 2(a_(1) b_(1) + a_(2) b_(2) + ... + a_(n) b_(n) x + (b_(1)^(2) + b_(2)^(2) + ... + b_(n)^(2) )` (2)
From (1), `f(x) ge0, AA x in` R . Hence, from (2), we have
` (a_1^(2) + a_(2)^(2) + ... + a_(n)^(2))x^(2) + 2(a_(1) b_(1) + a_(2) b_(2) + ... + a_(n) b_(n)) x + (b_(1)^(2) + b_(2)^(2) + ... + b_(n)^(2) )ge 0 AAx in`R
Discrimnant of tis corresponding equation is
D `le 0 (therefore` cofficient of `x^(2)` is positive )
`(a_(1)b_(1) + a_(2) b_(2) + ... + a_(n)b_(n))^(2) le (a_1^(2) + a_(2)^(2) + ... + a_(n)^(2)) (b_(1)^(2)+b_(2)^(2) + ... + b_(n)^(2))`
or `(a_(1)b_(1) + a_(2) b_(2) + ... + a_(n)b_(n))^(2) le (a_1^(2) + a_(2)^(2) + ... + a_(n)^(2)) (b_(1)^(2)+b_(2)^(2) + ... + b_(n)^(2))`