Prove ( a + b + c )^2 <= 4( ab + bc + ca )

1. Nov 10, 2006

Bernie Hunt

The Problem;
If a>0, b>0, c>0 and a + b > c, b + c > a, c + a > b then ( a + b + c )^2 <= 4( ab + bc + ca ).

Hints: Start with ( a + b + c )^2 and establish the inequality ( a + b + c )^2 <= 4( ab + bc + ca ). Us the inequality fact: if x < y then xy < y^2 if y > 0.

I haven’t make much progress. I’ve been trying;

( a + b + c )^2 = a^2 + b^2 + c^a + 2ab + 2bc + 2ca

But I haven’t been able to figure out a way to get the right side manipulated. I tried a couple of substitutions, but couldn’t justify a<b or such to be able to use the hint.

Can anyone give me a push in the right direction?

Thanks,
Bernie

Last edited: Nov 10, 2006
2. Nov 10, 2006

acm

I would start the question by expanding the RHS of the equasion to:
a^2 + b^2 + c^2 + 2(ab + bc + ac)

Now, It is obvious that (a + b + c)^2 >= 2(ab + bc + ac)
Hence 2(a + b + c)^2 >= 4(ab + bc + ac)
Therefore we can assume that 4(ab + bc + ac) = 2(a + b + c)^2
And as (a + b + c)^2 <= 2(a + b + c)^2
Therefore (a + b + c)^2 <= 4(ab + bc + ac)

3. Nov 11, 2006

Bernie Hunt

ACM,

Thanks for the quick reply. I think there is a flaw in your logic. The statements workout to;

2(a + b + c)^2 > (a + b + c)^2 > 2(ab + bc + ac)
2(a + b + c)^2 > 4(ab + bc + ac) > 2(ab + bc + ac)

Therefore 4(ab + bc + ac) > (a + b + c)^2

Which is saying that:
x > y
x > z
Therefore y > z

Which is not always true. It is true sometimes but not for all conditions.

Is there a way to prove that because a + b > c, b + c > a, c + a > b then the above logic is always true?

Bernie

4. Nov 11, 2006

StatusX

Expand out the square and cancel like terms. Then use their clue.

5. Nov 11, 2006

acm

No problem, Should have though that through.

6. Nov 11, 2006

matt grime

Either a=b and you can use that fact, or either a<b, or b<a and wlog we may assume that a<b in this case. (or, they're either all equal, when it is trivially true, or we may assume a<b without loss of generality - the labels of a,b,c are obviously not important).

7. Nov 11, 2006

Bernie Hunt

StatusX,

Cancel out which like terms? Expanding out the square gives;
( a + b + c )^2 = a^2 + b^2 + c^a + 2ab + 2bc + 2ca

Which of these terms can be canceled?

Bernie

8. Nov 11, 2006

StatusX

With the other side.

9. Nov 11, 2006

durt

And then rewrite 2ab as ab+ab, etc. Also, I don't think equality ever holds, unless you meant a + b >= c, b + c >= a, c + a >= b.

10. Nov 12, 2006

Bernie Hunt

OK, here we go;

Prove:
If a>0, b>0, c>0 and a + b > c, b + c > a, c + a > b then ( a + b + c )^2 <= 4( ab + bc + ca ).

Hints: Start with ( a + b + c )^2 and establish the inequality ( a + b + c )^2 <= 4( ab + bc + ca ). Us the inequality fact: if x < y then xy < y^2 if y > 0.

(a+b+c)^2 = (a+b+c)^2 (Reflective property of equality)

(a+b+c)^2 = a^2 + b^2 + c^a + 2ab + 2bc + 2ca

a + b > c
c(a+b) > c^2 fact of inequity
ca + cb > c^2

b + c > a
a(b+c) > a^2 fact of inequity
ab + ac > a^2

c + a > b
b(c+a) > b^2 fact of inequity
bc + ab > b^2

From above
(a+b+c)^2 = a^2 + b^2 + c^a + 2ab + 2bc + 2ca

By substitution
(a+b+c)^2 < (ab + ac) + (ca + cb) + (bc + ab) + 2ab + 2bc + 2ca

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

(a+b+c)^2 <= 4(ab + ac + bc) ???????????????????????

Can I make this last step from < to <=? If so is there a principal or property that justifies it?

Thanks,
Bernie

11. Nov 12, 2006

Think about what happens if a = b = c = 0.

12. Nov 12, 2006

StatusX

If something is less than something else, it is certainly also less than or equal to it. It's probably a typo though, and you were supposed to prove your next to last line.

13. Nov 12, 2006

Bernie Hunt

The proof is definately <=, but I think I found the justification. I don't have my notes handy, but I think it's disjunctive addtion justifies that if p then (p or q). Also if (p or q) and (~q) then p.

Bernie

14. Nov 12, 2006

StatusX

Yea, although it's pretty obvious if you just think about it. Also, I'm not saying you copied the question wrong, I'm saying it was wrong in the book. It's not strictly incorrect, but it's completely unnatural to use a $\leq$ in place of a < when equality is not possible.