Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$.

By Lagrange's identity $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2\color{red}{+(ad-bc)^2}.$$

Hint: use Cauchy-Schwarz in $\mathbb{R}^2$ on the vectors $(a,b)$ and $(c,d)$. This technique should provide a one-line proof of the desired result.

More directly, from what you've already computed, you can observe that $$ a^2d^2 - 2abcd + b^2c^2 = (ad - bc)^2 \geq 0 \text{,} $$ so $a^2d^2 + b^2c^2 \geq 2abcd$.

Diophantus has already shown that $$(a^{2}+b^{2})(c^{2}+d^{2})=(ac+bd)^{2}+(ad-bc)^{2}.$$ This proves the inequality.