# Inequality Proof

skims
Posers and Puzzles 01 Nov '08 22:52
1. 01 Nov '08 22:52
Given that 'a' and 'b' are unequal, positive whole numbers, can you prove that:

[a^3 + b^3] > [b(a^2) + a(b^2)] ?

I don't have the solution but would be interested in knowing the proof.

Thanks.
2. 01 Nov '08 23:24
show it first for the case b=a+1 and then procedd by induction on the case b=a+n
3. PBE6
Bananarama
02 Nov '08 06:42
Originally posted by skims
Given that 'a' and 'b' are unequal, positive whole numbers, can you prove that:

[a^3 + b^3] > [b(a^2) + a(b^2)] ?

I don't have the solution but would be interested in knowing the proof.

Thanks.
a^3 + b^3 > (a^2)b + a(b^2)

(a+b)(a^2 - ab + b^2) > (a+b)(ab)

Since a+b is strictly greater than 0, we can divide both sides by it:

a^2 - ab + b^2 > ab

a^2 - 2ab + b^2 > 0

(a-b)^2 > 0

Since a and b are distinct natural numbers, (a-b) must be an integer other than 0. Since the square of any integer other than 0 is always greater than 0, the given inequality must always be true.
4. 02 Nov '08 11:11
Rearrangement lemma. "It's just common sense"

If I have loads of cars, loads of cds, and loads of chocolate sprinkles, and you get to take 3 of one thing, 2 of another, and 1 of the third, what would you do?

3 cars, 2 cds, 1 chocolate sprinkle.

That's because to maximise the sum, you need to pair the largest items together (3 and car) and the smallest items together (1 and chocolate sprinkle).

Any other mixture, say, 2 cars, 3 chocolate sprinkles, 1 cd, will never be as big.

A and B are interchangable in the above formula, so we'll say A>B, and so A^2>B^2 cos they're whole and unequal.
We're matching up A^2 and B^2 with A and B.

Match biggest with biggest: A^2 with A
Match smallest with smallest: B^2 with B
and we get A^3+B^3 being bigger than any possible other mixture. In the equation, the mixture matches A^2 with B and B^2 with A.
5. ChronicLeaky
Don't Fear Me
02 Nov '08 14:211 edit
Originally posted by skims
Given that 'a' and 'b' are unequal, positive whole numbers, can you prove that:

[a^3 + b^3] > [b(a^2) + a(b^2)] ?

I don't have the solution but would be interested in knowing the proof.

Thanks.
Suppose there exist two distinct natural numbers a < b, such that

a^3 + b^3 < = ba^2 + ab^2.

Then:

a^2 (a - b) + b^2 (b - a) = (b-a)^2 (b + a) < = 0

The first factor is strictly positive, since a < b and the second is strictly positive since a,b > 0. Thus the last inequality is a contradiction and our assumption that such a and b exist must be incorrect, from which the inequality you want follows.

EDIT The idea is that one almost always establishes such inequalities by contradiction.