<s>[INST] Prove that $\forall a,b,c,d\in \mathbb{R}\,$, $(a^2-b^2)(c^2-d^2)\leq(ac-bd)^2$ and equalities hold if and only if $ad=bc$. Prove that $\forall a,b,c,d\in \mathbb{R}\,$,  $(a^2-b^2)(c^2-d^2)\leq(ac-bd)^2$ and equalities hold if and only if $ad=bc$.

I need hints in dealing with the inequality. I think what I have with dealing with the case of equality is sufficient. 
$\begin{align}(a^2-b^2)(c^2-d^2)\leq(ac-bd)^2&\Longleftrightarrow(ac)^2-(ad)^2-(bc)^2+(bd)^2\leq(ac)^2-2abcd+(bd)^2\\
&\Longleftrightarrow-(ad)^2-(bc)^2\leq-2abcd \\
\end{align}$
Suppose $ad\neq bc$, then $-(ad)^2-(bc)^2\leq-2abcd\Longrightarrow 2abcd\leq(ad)^2+(bc)^2,\,$ since $0<(ad)^2$ and $0<(bc)^2$ if $ad\neq bc$. Thus, the equality is only satisfied if and only if $ad=bc$, since this implies:
$$-(ad)^2-(bc)^2=-(ad)^2-(ad)^2=-2(ad)^2\leq-2abcd=-2ad\cdot ad=-2(ad)^2.$$
Hints to show inequality? Thanks.
 [/INST] You're almost there.
$$\
\begin{align}(a^2-b^2)(c^2-d^2)\leq(ac-bd)^2&\Longleftrightarrow(ac)^2-(ad)^2-(bc)^2+(bd)^2\leq(ac)^2-2abcd+(bd)^2\\
&\Longleftrightarrow-(ad)^2-(bc)^2\leq-2abcd \\
&\Longleftrightarrow 0 \le (ad)^2 - 2(ad)(bc) + (bc)^2\\
&\Longleftrightarrow 0 \le (ad - bc)^2\\
\end{align}$$
You get the $ad = bc$ condition proof for free here too.
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