Question

Challenge #2!

Let \(a,b,c \rm ~ be ~ distinct~ numbers ~\in R\).

The expressions given below are positive for all real \(x\) values.

\(ax^2 + bx + c\)
\(bx^2 + cx + a\)
\(cx^2 + ax + b\)

Let\[\alpha = \dfrac{ab + bc + ca}{a^2 + b^2 + c^2}\]Then:

A. \(\alpha > 1 \)
B. \(\alpha < 1\)
C. \(\alpha > 1/4\)
D. \(\alpha < 4\)

Check ALL THAT APPLY.

Answer 1

maybe i would start here :

Since the quadratics stay positive always, D < 0

b^2 -4ac < 0
c^2 - 4ab < 0
a^2 - 4bc < 0

Answer 2

Right way, @ganeshie8.

Answer 3

Adding :

b^2 -4ac < 0
c^2 - 4ab < 0
a^2 - 4bc < 0
---------------------
a^2 + b^2 + c^2 - 4(ab+bc+ca) < 0

Answer 4

Or : \(a^2 + b^2 + c^2 < 4(ab + bc + ca) \)

It would be easy now to conclude about alpha...

Answer 5

I think, it is C) \(\alpha > \cfrac{1}{4} \)

Answer 6

Yes, this is the easy part. But this is a more than one correct question.

Answer 7

Ohh

Answer 8

we need to find its upper bound also ?

Answer 9

It may or may not have an upper bound. That's why the question is confusing.

Answer 10

I kinda edited this question because it had a little hole. You may also want to think why a, b, c are distinct.

Answer 11

well for me instead of trying to reach the formula i can try to find which choice is correct

Answer 12

lol

Answer 13

Go ahead.

Answer 14

Using AM-GM may give us something.. hmm!

Answer 15

Yup !

Answer 16

lol Parth, can you do it from here? See, OpenStudy aims on providing the students a way to go through the question, we can not help you completely! You should try it first and let us know where you are getting problem!

Answer 17

Haha! :)

Answer 18

lulz

Answer 19

Should I give it out now?

Answer 20

Its up to you, we will not ask you to do so.. :P

Answer 21

clever way to escape typing the lengthy work @mathslover :P

Answer 22

Just wait a sec

Answer 23

lol , Thanks for the appreciation, it matters a lot for me , haha!

Answer 24

x^2(a + b + c) + x(a + b + c) + 1(a + b + c) > 0
(a + b + c)(x^2 + x + 1) > 0
x^2 + x + 1 > 0

Answer 25

\[a + b + c>0?\]

Answer 26

@Hero - Yes, its right but what will it give to us? x^2 + x + 1 will always be positive for all real numbers

Answer 27

And so, a + b + c > 0

Answer 28

Right.

Answer 29

I thought it would lead somewhere

Answer 30

Yeah but I am still trying to get a nice looking equation from that ! Till now - No good results :(

Answer 31

humm so b,c also cant be any numbers :\
lolz good luck solving it !

Answer 32

Oka, I'm implementing the lazy @mathslover idea about AM-GM to complete the problem :) :

a^2 + b^2 > 2ab
b^2 + c^2 > 2bc
c^2 + a^2 > 2ca
------------------
a^2 + b^2 + c^2 > ab+bc+ca

(ab+bc+ca)/(a^2+b^2+c^2) < 1a

Answer 33

YAY!

Answer 34

lol @ganeshie8 , that's so kind of you to call me lazy...

Answer 35

i think all the choices cud be true according to c,b u can get out with

Answer 36

Ganesh!

Ganesh!

Ganesh!

Ganesh!

Ganesh!

Answer 37

<dingdingdingding>

Alternative:

The answer follows from an equality we learnt in the 9th grade.\[a^2 + b^2 + c^2 - ab - bc - ca = \dfrac{1}{2}\left(( a- b)^2 + (b - c)^2 + (c-a)^2\right)\]But since \(a,b,c\) are distinct, we have \(\dfrac{1}{2}\left(( a- b)^2 + (b - c)^2 + (c-a)^2\right) > 0\).

In other words, \(a^2 + b^2 + c^2 - ab - bc - ca > 0\).

So...

Answer 38

Good work @ganeshie8 - My teachings worked finally :)

Answer 39

So, it will be B) And C) both... !

Answer 40

Yup.

Answer 41

Okay, so, challenge solved by @ganeshie8 ! ~~~~~~~~~~Closed

Answer 42

It was a nice time we had here, gentlemen.

Answer 43

Gentlemen? Who? I can't see any "gentle" men here? (except me..)

Answer 44

Funny.

Answer 45

Let's see what Unkle has got there for us.