Is there a process beyond just guessing and doing a lot of algebra to prove this inequality in Beckenbach's Inequalities book?

Question

There's a question in chapter 2 page 19 that says:

3. Show that $$ \left(a^2-b^2\right)^2 \geq 4 a b(a-b)^2 $$ for all $a, b$, and that the sign of equality holds if and only if $a=b$.

And the answer also included has no explanation:

3. Equivalent to $(a-b)^4 \geq 0$

And I completely believe it, but the book never presented any strategies for inequality solving so am I supposed to just expand everything out and then recognize that it looks like $(a-b)^4$ ? Is there a better way? Is there a different book I should be using if this one isn't explaining the tricks?

I'm just really bad at inequalities (especially with absolute values) and I want to get better at the basics so I can do well at Abbott's Analysis type problems

Answer 1

You can avoid expanding all of the brackets by spotting the common factor: $$\begin{aligned}(a^2-b^2)^2-4ab(a-b)^2&=(a-b)^2(a+b)^2-4ab(a-b)^2\\ &=(a-b)^2((a+b)^2-4ab)\\ &=(a-b)^2(a-b)^2\\ &=(a-b)^4 \end{aligned}$$

Answer 2

For $a=b$ there is evidently equality. You may assume that $a\neq b$. Write the left hand side as $$(a-b)^2(a+b)^2$$ Cancel the positive $(a-b)^2$ on both sides to get $$(a+b)^2>4ab$$ which is obvious, because it is equivalent to $$(a-b)^2>0$$ (remember that $a\neq b$).

This is not much of a trick, but also not much of algebra.