How do you prove the inequality (a^3+b^3)(a^2-ab+b^2) <= a^5+b^5?

[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...

 

[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?

 

[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.

 

[Dick]

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

 

[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.