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

  • Thread starter Bernie Hunt
  • Start date
In summary: This is a common issue with the substitution model, of course.In summary, we are trying to prove that 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 ). We can start by expanding out the square and establishing the inequality ( a + b + c )^2 <= 4( ab + bc + ca ) by using the fact that if x < y, then xy < y^2 if y > 0. By making substitutions and using the fact of inequality, we can show that ( a + b + c )^
  • #1
Bernie Hunt
19
0
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:
Physics news on Phys.org
  • #2
I would start the question by expanding the RHS of the equation 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
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
Expand out the square and cancel like terms. Then use their clue.
 
  • #5
No problem, Should have though that through.
 
  • #6
Bernie Hunt said:
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

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
StatusX,

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

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
With the other side.
 
  • #9
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
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
Think about what happens if a = b = c = 0.
 
  • #12
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
The proof is definitely <=, 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
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 [itex]\leq[/itex] in place of a < when equality is not possible.
 

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

1. What does the equation "Prove (a + b + c)^2 <= 4(ab + bc + ca)" mean?

This equation is known as the Cauchy-Schwarz inequality and is used in mathematics to prove the relationship between the sum of squares and the product of sums. It is often used in geometry and can also be applied to inequalities involving vectors.

2. How do you prove the Cauchy-Schwarz inequality?

There are multiple ways to prove this inequality, depending on the context and mathematical tools available. One common approach is to use the quadratic formula to derive the inequality from the given equation. Another method is to use the properties of dot products between vectors.

3. What are the applications of the Cauchy-Schwarz inequality?

The Cauchy-Schwarz inequality has various applications in mathematics, physics, and engineering. It is used in optimization problems, the study of inner product spaces, and in proving other inequalities. It also has applications in signal processing, statistics, and quantum mechanics.

4. Can the Cauchy-Schwarz inequality be generalized to more than three variables?

Yes, the inequality can be extended to any number of variables. The general form is (a1 + a2 + ... + an)^2 <= n(a1^2 + a2^2 + ... + an^2), where a1, a2, ... , an are real numbers. This is often referred to as the generalized Cauchy-Schwarz inequality.

5. How is the Cauchy-Schwarz inequality related to other mathematical inequalities?

The Cauchy-Schwarz inequality is closely related to other well-known inequalities, such as the Triangle Inequality and the Arithmetic Mean-Geometric Mean Inequality. It can also be used to prove other inequalities, such as the Rearrangement Inequality and the Hölder's Inequality.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
11
Views
455
  • Precalculus Mathematics Homework Help
Replies
4
Views
1K
  • Precalculus Mathematics Homework Help
Replies
5
Views
1K
  • General Math
Replies
1
Views
900
  • Math POTW for Secondary and High School Students
Replies
2
Views
954
  • Calculus and Beyond Homework Help
Replies
20
Views
2K
Replies
1
Views
859
Replies
1
Views
695
  • Precalculus Mathematics Homework Help
Replies
11
Views
1K
Back
Top