Inequality problem $(a^2-b^2)(a^4-b^4)\le (a^3-b^3)^2$ 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?

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.
