1. ali2x2

i saw tat question but i couldt help ;/

2. ali2x2

thats cheap asto

3. ganeshie8

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

4. ali2x2

ok :)

5. ParthKohli

inb4 Astrophysics tags me

6. ParthKohli

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

7. ganeshie8

Thats it!

8. 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 :)

9. ganeshie8

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

10. ParthKohli

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

11. anonymous

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

12. ganeshie8

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

13. ParthKohli

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

14. anonymous

no im not at that stage of mathematics yet

15. Astrophysics

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

16. ParthKohli

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

17. ganeshie8

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

18. Astrophysics

Yeah same, can you show it parth

19. Astrophysics

Oh I think you mean what op was already doing

20. ganeshie8

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

21. 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 : $\left(2\sum_{i=1}^{n}a_ib_i\right)^2-4\left( \sum_{i=1}^{n}a^2_i\right)\left( \sum_{i=1}^{n}b^2_i\right)$

22. ganeshie8

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

23. Astrophysics

perfect

24. Astrophysics

$\le 0$

25. ganeshie8

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

26. 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

27. 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!

28. Astrophysics

Haha, xD nice!

29. ali2x2

Why is this still up and viewing xD

30. ganeshie8

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