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?

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.
