A First Course in Mathematical Analysis, Ch 1, 1.6 Exercises, Section 1.3, ex 2 Prove that
\begin{align*}
(a^{2}+b^{2})(c^{2}+d^{2}) & \ge(ac+ad)^{2}\\
a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2} & \ge a^{2}c^{2}+2a^{2}cd+a^{2}d^{2}\\
b^{2}c^{2}+b^{2}d^{2} & \ge 2a^{2}cd\\
b^{2}c^{2}-2a^{2}cd+b^{2}d^{2} & \ge 0\\
b^{2}(c^{2}+d^{2})-2a^{2}cd & \ge 0\\
\end{align*}
Any hints on things to try? It seems trickier than say
$$
(a+b)^{2}(c+d)^{2}\ge(ac+bd)^{2}
$$
Proof (direct):
\begin{align*}
a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2} & \ge a^{2}c^{2}+2abcd+b^{2}d^{2}\\
a^{2}d^{2}+b^{2}c^{2} & \ge2abcd\\
a^{2}d^{2}-2abcd+b^{2}c^{2} & \ge0\\
a^{2}d^{2}-2(ad)(bc)+b^{2}c^{2} & \ge0\\
(ad-bc)^{2} & \ge0
\end{align*}

As written, the inequality is false.
$$(ac+ad)^2=a(c+d)^2=a^2c^2+2a^2cd+a^2d^2.$$
If $b=0 \text{ but } acd \neq 0 \text{ with } cd \gt 0$, then
$$(a^2+b^2)(c^2+d^2) = a^2(c^2+d^2) = a^2c^2+a^2d^2 \lt a^2c^2+2a^2cd+a^2d^2.$$
As noted in the comments, I suspect the right-hand side is supposed to be $(ac+\color{red} b d)^2.$
