Keeping the distance

Algebra Level 5

$\large\left|\frac{a_{1}b_{1} + a_{2}b_{2} + ... + a_{n}b_{n}}{\sqrt{a_{1}^2 + a_{2}^2 + ... + a_{n}^2}} \right| - \sqrt{ b_{1}^2 + b_{2}^2 + ... + b_{n}^2} > 0$ .

Let $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$ be real numbers for which the inequality above is fulfilled. Which of the answer choices given is true?

See my set of Problems .

a_{1} + a_{2} + \ldots+ a_{n} < -1

None of these choices

a_{1} + a_{2} +\ldots+ a_{n} < 1

a_{1} + a_{2} + \ldots+ a_{n} < 0

a_{1} + a_{2} + \ldots+ a_{n} < -2

a_{1} + a_{2} + \ldots + a_{n} < 2

2 solutions

Thushar Mn
Nov 23, 2015

|A.B|/|A| =|B| COS(t) ,so |B|COS(t)- |B| <0 ,always.....

Paul Ryan Longhas
Jul 10, 2015

Let $\vec{L} = ( a_{1}, a_{2}, ..., a_{n})$ and $\vec{M} = (b_{1}, b_{2}, ..., b_{n})$

Thus, $|\frac{\vec{L} \cdot \vec{M}}{||\vec{L}||}| - ||\vec{M}|| > 0$

So, $|\vec{L} \cdot \vec{M}| > ||\vec{L}||||\vec{M}||$ , which is a contradiction of Cauchy-Schwartz inequality.

Let each of b's value equals 0. Then, whatever value you put in any of a, the inequality comes up as, $\boldsymbol{\boxed{0-0>0}}$ , which cannot be true.

So, none of the choices are correct.

MD Omur Faruque - 5 years, 11 months ago

Keeping the distance

See my set of Problems .

2 solutions

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