Which is greater, a^4 + b^4 + c^4 or (ab)² + (ac)² + (bc)² - by how much?

Question

Assume that a, b and c are positive integers.
Questions like this is not very easy to answer - as far as my math knowledge is concerned. :)

Thanks Northstar and Yay!

I'm very interested some sort of special cases with some proof or details.

Let's say c = a + b as an example of a special case.

Thanks a lot Semyaza!
Even though you have not answered my original question, you have answered the case where a + b = c not only in integers but in real numbers!
Do you ever sleep man!? Lol!!

Thanks Northstar for updating your answer and by giving a very quick and neat answer - no offense to Semyaza who got into a deeper analysis.

Northstar and Semyaza, you're great men! As for you Go - I cannot comment. :)

Answer 1

Your problem can be restated as follows with little compromise on generalization:

Which is greater (aa)² + (bb)² + (cc)² or (ab)² + (ac)² + (bc)² where a, b and c are real numbers and |abc| > 0 and a + b + c = 0?

You are right when you say that this kind of problem is not relatively easy. We say “relatively” because this is elementary number theory stuff where students in B.S. Math are supposed to solve very quickly!

Okay let’s go straight to work.

Let ω = a² + b² + c² and a + b + c = 0.

Consider

(aa)² + (bb)²

= (aa - bb)² + 2aabb
= (a - b)²(a + b)² + 2aabb
= (a - b)²(a + b)² + 2aabb
= (a² - 2ab + b²)(a + b)² + 2aabb
= (a² - 2ab + b² + c² - c²)(a + b + c - c)² + 2aabb
= (ω - 2ab - c²)(0 - c)² + 2aabb
= (ω - 2ab - c²)(-c)² + 2aabb
= (ω - 2ab - c²)c² + 2aabb
= (ω - 2ab)c² - c²c² + 2aabb

Thus

(aa)² + (bb)² + (cc)² = (ω - 2ab)c² + 2aabb.

Now consider -c = a + b.

Squaring both sides gives us (-c)² = (a + b)² or c² = (a + b)².

Thus

c² = a² + 2ab + b²
2c² = a² + 2ab + b² + c² = ω + 2ab

This gives us

2c² - 2ab = ω or c² - ab = ½ ω.

Thus (aa)² + (bb)² + (cc)² = (ω - 2ab)c² + 2aabb reduces to

(aa)² + (bb)² + c²c²

= ωc² - 2abc² + 2aabb
= ωc² - 2ab(c² - ab)
= ωc² - 2ab(½ ω)
= ½ ω (2c² - 2ab)
= ½ ω (ω).

THUS, we have (aa)² + (bb)² + (cc)² = ½ ω² as Equation 1.

Now consider

(ab)² + (ac)² + (bc)²

= a²(b² + c²) + (bc)²
= a²(b² + c² + a² - a²) + (bc)²
= a²(ω - a²) + (bc)²
= a²ω - a²a² + (bc)²
= a²ω + (bc - aa)(bc + aa).

Recall that c² - ab = ½ ω.

Here if a + b + c = 0 then a² - bc = ½ ω as well.

Thus (ab)² + (ac)² + (bc)² = a²ω + (bc - aa)(bc + aa) reduces to

(ab)² + (ac)² + (bc)²

= a²ω + (-½ ω)(bc + aa)
= a²ω + (-½ ω)(bc + aa)
= ½ ω [2a² - (bc + aa)]
= ½ ω (a² - bc)
= ½ ω (½ ω)

THUS, we have (ab)² + (ac)² + (bc)² = ¼ ω² as Equation 2.

ERGO, IF a + b + c = 0 where a, b and c are real numbers and |abc| > 0 THEN (aa)² + (bb)² + (cc)² > (ab)² + (ac)² + (bc)² by ¼ (a² + b² + c²)².

So far this is NOT a general solution to your question BUT this is useful and comprehensive enough.
Have a nice day!

Answer 2

Obviously if a=b=c then the two are the same.

Otherwise assume a>b>c
then let a =b+x; b=c+y; giving a = c+x+y
a^4 = a^2(b+x)^2 = (ab)^2 + a^2(2bx+x^2)
b^4 = b^2(c+y)^2 = (bc)^2 + b^2(2cy +y^2)
c^4 = c^2(a-x-y)^2 = (ca)^2 + c^2(-2ax-2ay+2xy+x^2+y^2)

now, we just need to show that a^2(2bx+x^2) + b^2(2cy +y^2) + c^2(-2ax-2ay+2xy+x^2+y^2) is greater than 0 in order to prove the result.

I'll leave that up to the reader - just find a positive term obviously greater than each of the two negative terms!

Answer 3

lets say that a=1
b=2
and c =3
ok so the first one would be
1^4 which is 4
+ 2^4 which is 16
+3^4 which is 81 so these added up would be 91 now using the same numbers the other equation would be
(1*2)^2 which is 4
+ (1*3)^2 which is 9
+ (2*3)^2 which is 36 so now this added up is 49 so using those numbers the first equation will give you a greater amount

Answer 4

It depends on what the positive integers are.

Case 1) a = b = c
Then they are equal

a^4 + b^4 + c^4 = (ab)² + (ac)² + (bc)²
a^4 + a^4 + a^4 = (a²)² + (a²)² + (a²)²
3a^4 = 3a^4

Case 2) They are not all equal.
Then

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

Special Case

c = a + b

Then

a^4 + b^4 + c^4 = a^4 + b^4 + (a + b)^4
= 2a^4 + 4a³b + 6a²b² + 4ab³ + 2b^4

(ab)² + (ac)² + (bc)² = (ab)² + [a(a + b)]² + [b(a + b)]²
= a^4 + 2a³b + 3a²b² + 2ab³ + 1b^4

So we have:

a^4 + b^4 + c^4 = 2{(ab)² + (ac)² + (bc)²}

The left hand side is twice as great.

Answer 5

If you are trying to get someone to do your homework you have come to the wrong place. And to all the parents and teachers that lie to their kids and their students and tell them they are going to use information like this in reality, you are all full of s__t, and we never had time for it.

And to all the people who give my answer a thumbs down, you are only trying to insult my answer because you know I am right.

Answer 6

I can follow and approve Semyaza's answer but it's a sad thing he doesn't provide a case where a + b + c ≠ 0!
Can you please do me a good favor? Please try to have a solution for a + b + c ≠ 0 which I myself is very interested in discovering. In behalf of John, Thanks in advance!