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This problem that I have been unable to solve is from the book "Introduction to Inequalities" by Beckenbach and Bellman, chapter 2, page 22, problem 4.

Problem 4. Show that $$(a^2-b^2)(a^4-b^4)\le (a^3-b^3)^2$$ and $$(a^2+b^2)(a^4+b^4)\ge (a^3+b^3)^2$$ for all $a,b$.

My try for first part of the problem, just so you know that I have done some work, is this:

We have $$(a^2-b^2)(a^4-b^4)\le (a^3-b^3)^2$$ by factoring out, we are left with $$(a-b)(a+b)(a-b)(a+b)(a^2+b^2)\le (a^3-b^3)^2,$$ $$(a-b)^2(a+b)^2(a^2+b^2)\le (a-b)^2(a^2+ab+b^2)^2,$$ $$a^4+2a^2b^2+2a^3b+2ab^3+b^4\le a^4+2a^3b+ab^3+2a^2b^2+2ab^3+b^4,$$we get $$ab^2\ge 0$$which is true but for only $a,b\ge 0$, so my solution is false, can anyone hint me to correct way of solving this?

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You've written $a^2+ab+b^2$ where you meant its square, and then I think you've miscalculated the square. Anyway, the symmetry in the powers of $a$ and $b$ makes $ab^2\ge0$ an unlikely thing to get. Try the algebra again. –  Gerry Myerson Feb 9 '13 at 11:40
A comment on your working: Why did $(a^3 - b^3)^2$ change to $(a^3 - b^3)$ and then change to $(a - b)^2 (a^2 + ab + b^2)$. It should be $(a^3 - b^3)^2 = (a - b)^2 (a^2 + ab + b^2)^2$. –  muzzlator Feb 9 '13 at 11:40
yes, I am sorry, I misspelled that. –  Paul Dirac Feb 9 '13 at 11:43
I forgot squares here, but not on paper, so calculations I think are correct. –  Paul Dirac Feb 9 '13 at 11:44
No, you have miscalculated the square of $a^2+ab+b^2$, and anyway your last inequality doesn't follow from the one before it --- where is there an $ab^2$ in the next-to-last inequality? Try the algebra again, you're almost there. –  Gerry Myerson Feb 10 '13 at 9:34

1 Answer 1

up vote 3 down vote accepted

If $a = 0$ or $b = 0$, the two sides are equal, and hence the inequality holds. Now, let us assume $a \neq 0$ and $b \neq 0$:

$(a^2 - b^2)(a^4 - b^4) \leq (a^3 - b^3)^2$
iff $a^6 + b^6 - a^2b^2(a^2 + b^2) \leq a^6 + b^6 - 2a^3b^3$
iff $- a^2b^2(a^2 + b^2) \leq - 2a^3b^3$
iff $a^2b^2(a^2 + b^2) \geq 2a^3b^3$
iff $(a^2 + b^2) \geq 2ab\ \ \ $ ($a, b \neq 0$)
iff $a^2 + b^2 - 2ab \geq 0$
iff $(a - b)^2 \geq 0$
which is always true.

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I went down that root several times but I missed something it seems, thank you very much for your help! +1. –  Paul Dirac Feb 9 '13 at 11:48

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