Question

Given that a,b and c are possible real numbers,such that a+b+c = x,then which of the following is true?
1) (x-a)(x-b)(x-c) >= 8abc
2) 1/a + 1/b + 1/c >= 9/x
3) (x-a)(x-b)(x-c) >= 8/27a^3

Answer 1

possible real numbers?

Answer 2

is that right?

Answer 3

yes a,b,c are possible real numbers

Answer 4

im just confused why it would say possible instead of saying let a,b,c be complex numbers

Answer 5

a,b and c can be any real number its not a complex number for sure.

Answer 6

Yeah, it sounds like it means arbitrary real number, not that the number might be real.

Answer 7

well its possible for a,b, and c to be real
but there is also the possibility of it being complex
we arent sure what a b and c are

Answer 8

so it doesn't say positive right?

Answer 9

yes there can be possibility of a,b,c being complex numbers.but the question ask for the solution considering a,b,c being real and positive only

Answer 10

i don't see positive anywhere
i see possible which means we aren't sure what a b and c are

Answer 11

so you are saying we get to assume a, b, and c are postive real numbers

Answer 12

yes.

Answer 13

1

Answer 14

You would have to assume that otherwise you get contradictions.

Answer 15

can you please let me know how you came to conlusion

Answer 16

how did you find the solution.???

Answer 17

The first one does seem to work out.

Answer 18

consider a=1,b=2,c=3 => x=6 ; option 1 and 2 both will get satisfied

Answer 19

Yes, the second one seems to work as well. Both work for positive and negative test I've tried. Haven't tried the third, but there must be a more rigorous way of solving this.

Answer 20

yup...!

Answer 21

What you want to do is eliminate the incorrect solutions by coming up with a,b,c that makes the inequality fail. For the third one, a = 1, b = 0, c= 0 makes x = 1, so the left side is 0, the other is 8/27, so that one fails.

For the second one, try a =1, b = -1, c = 2. Then 1/a+1/b+1/c = 1/2, but 9/x = 9/2, which is bigger, so that fails too. This leaves only the first one as a possibility.

Answer 22

(x-a)(x-b)(x-c)
=(x^2-ax-bx+ab)(x-c)
=x^3-ax^2-bx^2+abx-x^2x+axc+bxc-abc
=x^3-x^2(a+b+c)+x(ab+ac+bc)-abc
=x^3-x^2*x+x(ab+ac+bc)-abc
=x^3-x^3+x(ab+ac+bc)-abc
=0+(a+b+c)(ab+ac+bc)-abc
=(a+b+c)(ab+ac+bc)-abc
=a^2b+a^2c+abc+ab^2+abc+b^2c+abc+ac^2+bc^2    -abc
=2abc+a^2(b+c)+b^2(a+c)+c^2(a+b)
=2abc+a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)
>2abc+a(b+c)^2+b(a+c)^2+c(a+b)^2
=2abc+a(b^2+2cb+c^2)+b(a^2+2ac+c^2)+ c(a^2+2ab+b^2)
=8abc+a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)
>8abc
assuming
a>0,b>0,c>0

Answer 23

that a great way of solving..!like it

Answer 24

<2abc+a(b+c)^2+b(a+c)^2+c(a+b)^2*
i made a mistake up there
so this changes things
:(

Answer 25

none of the option will get satisied if you will take a=b=c=1