# Another inequality proof

1. Apr 16, 2007

### siddharthmishra19

The problem

For all a>0 and b>0 if a^3+b^3 = a^5+b^5 prove that
a^2+b^2<=1+ab

I have no idea on even how to start... i have tried using trigonometry (like in previous post) but come to a dead end... i am looking for the simplest method...

2. Apr 16, 2007

### Dick

The only proof I've been able to come up with is kind of weird. Maybe you can do better. You should easily be able to convince yourself that if both a and b are >1 the equality can't hold. Same for <1. So we can take a>=1 and b<=1. Now consider f(n)=a^n+b^n (n>0). Now show f(n) has one extermum and it's a minimum. Your premise says f(3)=f(5). This says the extremal n is between 3 and 5. So f(1)>=f(3). Or a+b>=a^3+b^3. Factor the LHS and divide by (a+b) and you have your result. Funny, huh?

3. Apr 17, 2007

### drpizza

I tried starting with a different approach. Since LHS = RHS, then RHS/LHS = 1. Doing a little creative long division, I obtained:

$$\frac{a^5+b^5}{a^3+b^3} = 1 = a^2 + b^2 - \frac{a^3 b^2 + a^2 b^3}{a^3 + b^3}$$

From there,
$$1 + \frac{a^3 b^2 + a^2 b^3}{a^3 + b^3} = a^2 + b^2$$

Then, perhaps on the LHS, factor out an ab:
$$1 + ab( \frac{a^2 b + a b^2}{a^3 + b^3}) = a^2 + b^2$$

I'd think you can turn it around to $$a^2 + b^2 = 1 + ab( \frac{a^2 b + a b^2}{a^3 + b^3}) = (re-written) <= re-written with a term dropped out.$$

I'm drawing a momentary blank (and have to get home!) but I can't "see" the next step from here; but maybe it's enough that you can continue.

Last edited: Apr 17, 2007
4. Apr 17, 2007

### Dick

Keep trying, drpizza. I'd love to see the non-calc proof. I tried stuff like that for quite a while.

5. Apr 18, 2007

### AKG

It's equivalent to show that:

$$(a^3 + b^3)(a^2 - ab + b^2) \leq a^5 + b^5$$

This line is true iff:

$$a^5 + a^2b^3 - a^4b - ab^4 + a^3b^2 + b^5 \leq a^5 + b^5$$

iff

$$a^2b^3 - a^4b - ab^4 + a^3b^2 \leq 0$$

iff

$$ab^2 - a^3 - b^3 + a^2b \leq 0$$

iff

$$a(b^2 - a^2) - b(b^2 - a^2) \leq 0$$

iff

$$(a-b)(b^2 - a^2) \leq 0$$

iff

$$(a-b)(b-a)(b+a) \leq 0$$

iff

$$-(a-b)(a-b)(b+a) \leq 0$$

iff

$$(a-b)^2(b+a) \geq 0$$

Well (a-b)2 is a square, hence non-negative. b and a are both positive, so (b+a) is positive. So the product on the left is indeed non-negative, so the desired result holds.