http://assets.openstudy.com/updates/attachments/55c4a0c1e4b0c7f4a978d9c7-praxer-1438949589754-screenshot_58.png

- ganeshie8

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- katieb

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- ali2x2

i saw tat question but i couldt help ;/

- ali2x2

thats cheap asto

- ganeshie8

there is really a very simple and cute solution, give it a try again :)

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## More answers

- ali2x2

ok :)

- ParthKohli

inb4 Astrophysics tags me

- ParthKohli

These are perfect squares. The only way we have it is that each one is zero.

- ganeshie8

Thats it!

- ali2x2

I couldnt find a way but now that i see the correct answer it clears up the fog in my mind, good job @ParthKohli :)

- ganeshie8

please finish it off @ParthKohli
i think @praxer is still looking for a solution

- ParthKohli

Oh, sorry. OS is acting up for me again. I type out things and it removes them.

- anonymous

this suddenly look like ultimate troll question after you explain it lol

- ganeshie8

Haha if you don't like the simplicity, there is a complicated solution, which is quite enlightening too :)
Familiar with Cauchy-Schwarz inequality ?

- ParthKohli

IT REMOVED IT AGAIN!
Yes, I actually solved the same question on my test.

- anonymous

no im not at that stage of mathematics yet

- Astrophysics

\[\sum_{i=1}^{n} (a_ix+b_i)^2 = x^2 \sum_{i=1}^{n}a^2_i+2x \sum_{i=1}^{n}a_ib_i+\sum_{i=1}^{n}b_i^2=0\]

- ParthKohli

The OP isn't actually too far from directly solving it. You can just expand and complete the square.

- ganeshie8

what do you mean by directly expand and complete the square ?
i thought we will have to use cauchy-schwarz inequality

- Astrophysics

Yeah same, can you show it parth

- Astrophysics

Oh I think you mean what op was already doing

- ganeshie8

Ohkiee, I'll finish off your solution using geometry :
|dw:1438963298665:dw|

- ganeshie8

\[\sum_{i=1}^{n} (a_ix+b_i)^2 = x^2 \sum_{i=1}^{n}a^2_i+2x \sum_{i=1}^{n}a_ib_i+\sum_{i=1}^{n}b_i^2=0\]
then discriminant is given by :
\[\left(2\sum_{i=1}^{n}a_ib_i\right)^2-4\left( \sum_{i=1}^{n}a^2_i\right)\left( \sum_{i=1}^{n}b^2_i\right) \]

- ganeshie8

thats what OP had too, so we're on same page :)

- Astrophysics

perfect

- Astrophysics

\[\le 0\]

- ganeshie8

Ahh right, but whats the justification for \(\le 0\) ?

- Astrophysics

Well I think since \[\sum_{i=1}^{n} (a_ix+b_i)^2 = 0\] is only way we can get real roots, so each term in the sum will have to be 0, and I don't really know the exact proof of it goes, but if we want what you have above \[\left(2\sum_{i=1}^{n}a_ib_i\right)^2-4\left( \sum_{i=1}^{n}a^2_i\right)\left( \sum_{i=1}^{n}b^2_i\right)\] this would need to be <= 0, for the roots of \[x^2 \sum_{i=1}^{n}a^2_i+2x \sum_{i=1}^{n}a_ib_i+\sum_{i=1}^{n}b_i^2=0\] exist

- ganeshie8

That looks good! since the quadratic \(\sum_{i=1}^{n} (a_ix+b_i)^2 \) is nonnegative, its determinant must be \(\le 0\) :
\[\left(2\sum_{i=1}^{n}a_ib_i\right)^2-4\left( \sum_{i=1}^{n}a^2_i\right)\left( \sum_{i=1}^{n}b^2_i\right) \le 0\]
We just proved Cauchy-Schwarz inequality!

- Astrophysics

Haha, xD nice!

- ali2x2

Why is this still up and viewing xD

- ganeshie8

one of the cool things in math in general is TMTOWTDI ;p

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