# A problem

1. Dec 10, 2004

### Anzas

there are three numbers a,b,c
0<a,b,c<=1
and it is given that
ab+ac+bc=1

show that
a^2+b^2+c^2 <=2

2. Dec 10, 2004

### dextercioby

I believe there's an error with your sign:it should be:
$$a^{2}+b^{2}+c^{2} \geq 2$$

Use the square formula and the fact that the square of a real number is always greater than "0".

3. Dec 11, 2004

### matt grime

I don't think so, some how. since there is obviously a case where a=b=c and thus a^2+b^2+c^2=1.

4. Dec 11, 2004

### dextercioby

Then what am I doing wrong??
$$(a-b)^2 \geq 0$$ => $$a^2+b^2 \geq 2ab$$ (1).
$$(a-c)^2 \geq 0$$ => $$a^2+c^2 \geq 2ac$$ (2).
$$(b-c)^2 \geq 0$$ => $$b^2+c^2 \geq 2bc$$ (3).

Add all 3 relations,devide by 2 and u'll be left with:
$$a^2+b^2+c^2 \geq 1$$ (4).

Which is different than i had previously obtained last night (i was really tired) but it's not what the problem's asking.

5. Dec 11, 2004

### arildno

The original problem seems a bit tricky.
I've been playing with one idea as how to prove it:
1) Define vectors (a,b,c),(b,c,a).
Then, by our given equality, written as a dot product:
$$(a,b,c)\cdot(b,c,a)=(a^{2}+b^{2}+c^{2})\cos\theta=1$$
Or:
$$a^{2}+b^{2}+c^{2}=\frac{1}{\cos\theta}$$
If I therefore could prove $$\cos\theta\geq\frac{1}{2}$$
I would have solved it..
EDIT
Oops, the original problem should probably be solved using Lagrange multipliers.

Last edited: Dec 11, 2004
6. Dec 11, 2004

### philosophking

Wow, I really don't think calc3 is necessary (also, wouldn't you need stricter conditions on a,b,c other than they're between 0 and 1?). I also, now that I'm thinking about it, am not entirely sure how to solve it. This was my original idea:

0<=(a+b+c)^2
0<=a^2+b^2+c^2+2ab+2ac+2bc
0<=a^2+b^2+c^2 (this is a given like the original I think?)
and
2ab+2bc+2ac=2(1)=2
so
0<=a^2+b^2+c^2 +2
but... *sigh* Maybe someone else knows where one could take it from here? Could we put better restrictions on a,b,c?

7. Dec 11, 2004

### dextercioby

No,your put your mind at work only to come up with nothing.U just proved that the sum of the squares is larger than -2,which is more than obvious,since a sum of (real number) squares is always >=0>-2.

8. Dec 11, 2004

### gonzo

Well, can someone show that:

(a-1)^2 <= bc

Cause from there it's easy ...

9. Dec 11, 2004

### gonzo

I thought I had it, but realized at the end I had a small error. Here are some things I did find that seem like they were leading in the right direction (until I got too tired).

You can start with a>= 1/2 which gives 2a>=1 and maybe use that somehow (this is easy to show).

If you can get to a+b+c <= 2 the end result is also easy, this seemed like an easier initial target.

10. Dec 12, 2004

### gonzo

Okay, I think I've got it. A kind of weird proof, so let me know if anyone sees a flaw.

First we order the letters in an arbitrary order:

0 < c <= b <= a <= 1

-3a^2 <= 0
2a^2 - 4a^2 - a^2 <= 0

since c<=a:
2a^2 - 4ac - c^2 <= 0
2a^2 + c^2 <= 4ac + 2c^2

sinc b<=a:
a^2 + b^2 + c^2 <= 4ac + 2c^2
a^2 + b^2 + c^2 <= 2ac + 2ac + 2c^2

sinc b>=c:
a^2 + b^2 + c^2 <= 2ac + 2ab + 2bc

so:

a^2 + b^2 + c^2 <= 2

11. Dec 12, 2004

### dextercioby

There's one small problem with this step:
$$2a^2-4a^2-a^2 \leq 0$$
U assume that $c\leq a$,but my guess is that u cannot full justify that:
$$2a^2 \leq 4ac+c^2$$
,because,as a,b,c are arbitray in the domain (0,1),while 'c'(which u assumed to be the smallest of the 3) could be very small,close to 0,which would mean that your relation would fail,right??

Last edited: Dec 12, 2004
12. Dec 12, 2004

### gonzo

I had a sign error ... I'll have to look it over again, but I think I can do it in a similar way getting the relations right.

13. Dec 12, 2004

### matt grime

Nothing (it's simply the AM-GM inequality) however this isn't what you said in the previous post is it, as you point out.

14. Dec 13, 2004

### gonzo

Okay, now I think I have it. Once again, start with an arbitrary ordering:

1>=a>=b>=c>0

so:
a-c <= 1
a(a-c) <= 1
a^2 - ac <= 1
a^2 + b^2 +c^2 - b^2 - c^2 -ac <= 1
a^2 + b^2 +c^2 - ab - bc -ac <= 1
a^2 + b^2 +c^2 <= 2

Let me know if anyone sees a problem with this. I think I got all the signs right this time.

15. Dec 14, 2004

### gonzo

Isn't anyone going to tell me if they think I'm right or wrong?

16. Dec 14, 2004

### matt grime

I think you're right, though I'd write it as:

wlog a=max(a,b,c)

a^2+b^2+c^2 <= a^2+ ab+ac = a^2 + 1-ac <= 1+a^2 <=2.

17. Dec 14, 2004

### Rogerio

Very nice, Matt !

:-)

18. Dec 14, 2004

### MathematicalPhysicist

in your second line you multiplied the left side of the inequality with 'a' but forgot to multiply the right side of the inequality with 'a'.
another thing if you are going with your first assumption:
1>=a>=b>=c>0
then when you reduct c from everything you get:
1-c>=a-c
and not a-c<=1.

19. Dec 14, 2004

### shmoe

You suggest a(a-c)<=a, but since a<=1 also, the weaker inequality a(a-c)<=1 is also true and sufficient for the problem. Same thing for your other complaint, a-c<=1-c implies a-c<=1 since c is positive.

20. Dec 14, 2004

### MathematicalPhysicist

if you assume a=1 then it's obviously right.

i would think to solve this question you simply need to prove that a+b+c<=2
because then when squaring both sides you get (a+b+c)^2<=4 and substracting 2 from both sides get you with a^2+b^2+c^2<=2, and if he does assume a=1 then the other numbers b,c are fractions which their product should with their sum be equal to one which means that b+c is indeed smaller than one and therefore this a+b+c<=2 is correct and so what that deduced from this.

btw, this question was asked in another forum (israeli one), and it's fair from the poster who post this question not even to say from where it was taken. :grumpy: