Prove a^2+b^2+c^2 ≥ ab + ac + bc | ABC

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In summary, the conversation discusses a proof of the identity a^2+b^2+c^2>=ab+ac+bc and different approaches to proving it, including using the identity a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}[(a-b)^2+(b-c)^2+(c-a)^2] and considering special cases. It also includes some humorous comments and a mention of missing dinosaurs.
  • #1
abc
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can anyone prove this :
a^2+b^2+c^2 ( is greater or equal to ) ab + ac + bc
thanx
regards
abc
 
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  • #2
A very important identity in mathematics is:
1+1=2
Note that this is readily transformed into another important identity:
[tex]\frac{1}{2}+\frac{1}{2}=1[/tex]
now, try to rewrite
[tex](\frac{1}{2}+\frac{1}{2})a^{2}+(\frac{1}{2}+\frac{1}{2})b^{2}+(\frac{1}{2}+\frac{1}{2})c^{2}-ab-ac-bc[/tex]
 
  • #3
This is like...
a^2+b^2 >= ab, right?

take (a-b)^2 >= 0.
(a-b)^2=a^2+b^2-2ab.
so a^2+b^2 >= 2ab >=ab.

so now u want a^2+b^2+c^2,
so take (a-b-c)^2, and (a+b-c)^2, etc...
and do the same sort of thing, a bit trickier though.
 
  • #4
arildno's approach is much nicer...but forgive his efforts at humor.
 
  • #5
**Sigh**

I miss the dinosaurs. -ss
:rofl:
 
  • #6
How about...we assume a^2 + b^2 + c^2 < ab + ac + bc and consider a special case where 'a' being minus and absolute value of 'a' is greater than the absolute value of b for ovious reason to lead a contradiction. So it will prove the negation of what assumed is true.
 
  • #7
it will only provide a contradiction in the special case where a is negative and in abs value greater then b, which need not be true.
 
  • #8
Use the identity for [tex]a^2 + b^2 + c^2 - ab - bc - ca[/tex]:

[tex]
a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}[(a-b)^{2} + (b-c)^{2} + (c-a)^{2}]
[/tex]

The right hand side is always greater than or equal to zero (equality in the case a = b = c). This proves the result.

Hope that helps...

Cheers
Vivek
 
Last edited:
  • #9
maverick280857 said:
Use the identity for [tex]a^2 + b^2 + c^2 - ab - bc - ca[/tex]:

[tex]
a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}[(a-b)^{2} + (b-c)^{2} + (c-a)^{2}]
[/tex]

The right hand side is always greater than or equal to zero (equality in the case a = b = c). This proves the result.

Hope that helps...

Cheers
Vivek
This is exactly what arildno was saying - without actually putting the spoon in the mouth.
 
  • #10
Oh well I didn't quite figure that out and since the question seemed unanswered to me so I went ahead and posted the solution (put the spoon in the mouth if you like it that way) :-). Its been quite a while since it was posted anyway.

Cheers
Vivek
 

1. What does the inequality a^2+b^2+c^2 ≥ ab + ac + bc | ABC mean?

The inequality a^2+b^2+c^2 ≥ ab + ac + bc | ABC means that the sum of the squares of three numbers (a, b, and c) is greater than or equal to the sum of all possible combinations of two of those numbers multiplied together (ab, ac, and bc). This notation is often used in mathematics to represent a relationship between multiple variables.

2. How do you prove the inequality a^2+b^2+c^2 ≥ ab + ac + bc | ABC?

The inequality a^2+b^2+c^2 ≥ ab + ac + bc | ABC can be proven using various mathematical techniques such as algebra, geometry, or calculus. One approach could be to rearrange the terms and then use the properties of inequalities to simplify the equation and reach a conclusion. Another approach could be to use geometric reasoning or visual representations to demonstrate the relationship between the terms.

3. What is the significance of the inequality a^2+b^2+c^2 ≥ ab + ac + bc | ABC?

The inequality a^2+b^2+c^2 ≥ ab + ac + bc | ABC is significant because it represents a fundamental relationship between the sum of squares and the sum of products of three numbers. This relationship has many applications in mathematics, physics, and engineering, and is often used to solve problems involving optimization, inequalities, and number theory.

4. Can the inequality a^2+b^2+c^2 ≥ ab + ac + bc | ABC be reversed?

No, the inequality a^2+b^2+c^2 ≥ ab + ac + bc | ABC cannot be reversed. This is because reversing the inequality would result in a false statement, as the sum of squares is always greater than or equal to the sum of products of three numbers. However, the inequality can be made strict by removing the equal sign, i.e. a^2+b^2+c^2 > ab + ac + bc | ABC, in which case it can be reversed to get a < b < c.

5. How is the inequality a^2+b^2+c^2 ≥ ab + ac + bc | ABC related to the Pythagorean theorem?

The inequality a^2+b^2+c^2 ≥ ab + ac + bc | ABC is closely related to the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (c) is equal to the sum of the squares of the other two sides (a and b). By substituting c with the sum of a and b, we get a^2+b^2+c^2 = ab + ac + bc. This shows that the Pythagorean theorem is a special case of the inequality, where all three numbers are positive and equal.

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