Proving Inequality: a^2+b^2+c^2 <= 2 for 0<a,b,c<=1 and ab+ac+bc=1

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The discussion revolves around proving the inequality a^2 + b^2 + c^2 ≤ 2 under the conditions 0 < a, b, c ≤ 1 and ab + ac + bc = 1. Participants explore various mathematical approaches, including the use of the square formula and vector definitions, while debating the correctness of their assumptions and calculations. Some suggest using Lagrange multipliers or the AM-GM inequality as potential methods for proof. A consensus emerges that establishing a + b + c ≤ 2 is crucial for proving the original inequality. The conversation highlights the complexity of the problem and the need for careful consideration of conditions and inequalities.
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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
 
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Anzas said:
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


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".
 
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.
 
matt grime said:
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.

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.
 
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.
 
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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?
 
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.
 
Well, can someone show that:

(a-1)^2 <= bc

Cause from there it's easy ...
 
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
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
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??
:confused:
 
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  • #12
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
dextercioby said:
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.

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
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
Isn't anyone going to tell me if they think I'm right or wrong?
 
  • #16
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
matt grime said:
wlog a=max(a,b,c)

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

Very nice, Matt !

:-)
 
  • #18
gonzo said:
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.
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
loop quantum gravity said:
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.

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
shmoe said:
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.
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.
 
  • #21
loop quantum gravity said:
if you assume a=1 then it's obviously right.

There's no need to assume a=1, his inequalities are still correct.
 
  • #22
loop quantum gravity said:
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.

I think you are a bit off here. First, I didn't multiply both sides of the inequality by "a", though as was pointed out that would be fine too. If you multiply two numbers between 0 and 1 you end up with another number between 0 and 1.

And if a>=c and a and c are between 0 and 1 then of course a-c<=1.

matt grime said:
wlog a=max(a,b,c)

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

Okay, much much much nicer than mine ;)

Two questions though.

what is "wlog"?
and I assume you mean 1-bc and not 1-ac (small nit)?
 
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  • #23
Ok, 1-bc then.

wlog means without loss of generality (ie a made a the largest just for ease, i could have picked one of the other ones to be the largest, and the argument is the same up to relabelling, etc)
 
  • #24
Cool, I thought it was some strange kind of logarithm. I'll have to remember that in the future.
 
  • #25
gonzo said:
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.

Wow! I was staring at this problem for ages and couldn't solve it. Very nice gonzo!
 

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