Showing that $(a^2 - b^2)^2$ $ \ge $ $4ab(a-$ $b)^2$ An inequality problem from Beckenbach and Bellman:

Show that $(a^2 - b^2)^2 \ge 4ab(a-b)^2$

The given answer is simply

Equivalent to $(a - b)^4 \ge 0$

I have tried two approaches, one which agrees with the given answer, and the other which does not.
Approach one. (Agrees with answer)
\begin{align} 
(a^2 - b^2)^2 & \ge 4ab(a-b)^2\\ 
(a^2 - b^2)^2 - 4ab(a-b)^2 & \ge 0\\  
((a+b)(a-b))^2 - 4ab(a-b)^2 & \ge 0\\
(a+b)^2(a-b)^2 - 4ab(a-b)^2 & \ge 0\\
(a-b)^2((a+b)^2 - 4ab) & \ge 0 \\
(a-b)^2 (a^2 -2ab + b^2) &\ge 0 \\
(a-b)^2 (a-b)^2 & \ge 0\\
(a - b)^4 & \ge 0
 \end{align}
Approach Two
\begin{align}
(a^2 - b^2)^2 & \ge 4ab(a-b)^2\\
((a+b)(a-b))^2 & \ge 4ab(a-b)^2\\
(a+b)^2(a-b)^2 & \ge 4ab(a-b)^2\\
(a+b)^2 & \ge 4ab\\
(a^2 -2ab + b^2) &\ge 0 \\
(a-b)^2 & \ge 0
\end{align}
Could someone point out where the second approach is going wrong?
 A: Let's examine your second attempt in more detail and check to see what's going wrong with it.
Observation 1
Whenever we are dividing using algebraic terms, it is vital that we make sure that we are not dividing by zero. Otherwise, the proof can end up going very wrong. There are many examples of this going wrong online including an incorrect proof that $0=1$ which is only possible because of the fact that the proof involves a division by a term equal to zero.
Therefore, when you divide by $(a-b)^2$ we want to consider the case where $a=b$ separately for a complete proof.
Observation 2
The inequality $(a-b)^2\ge 0$ which find at the end of your second proof is equivalent to the result you are looking for.
Note: since $(a-b)^2$ is non-negative we can square both sides of the inequality without reversing the "$\ge$" symbol. This gives us the desired inequality $(a-b)^4 \ge 0$.
Conclusion
Therefore, there is no problem with your second proof for the case where $a \neq b$.
However, when $a=b$, we must start again from the first inequality and replace every instance of $b$ with $a$ (or vice versa) to give the inequality:
$(a^2 - a^2)^2 \ge 4a^2(a-a)^2$ $\iff$ $0 \ge 0$ which is clearly true.
As we have shown that the inequalities are equivalent for $a=b$ and $a\neq b$, this completes the proof.
A: Note that
$$
a^2 - b^2 = (a + b) (a - b)
$$
Thus,
$$
(a^2 - b^2)^2 = (a + b)^2 (a - b)^2
$$
Thus, the given inequality
$$
(a^2 - b^2)^2 \geq 4 a b (a - b)^2
$$
is equivalent to
$$
(a + b)^2 (a - b)^2 \geq 4 a b (a - b)^2
$$
or
$$
(a - b)^2 \left[ (a + b)^2 - 4 a b \right] \geq 0
$$
or
$$
(a - b)^2 \left[ a^2 + b^2 - 2 a b \right] \geq 0
$$
or
$$
(a - b)^2 (a - b)^2 \geq 0
$$
or
$$
(a - b)^4 \geq 0
$$
which is always true.
Hence, we proved the given inequality.
