TITLE: Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$.
QUESTION [7 upvotes]: Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$.
So $(a^2+b^2)(c^2+d^2) = a^2c^2+a^2d^2+b^2c^2+b^2d^2$
and $(ac+bd)^2 = a^2c^2+2acbd+b^2d^2$
So the problem is reduced to proving that $a^2d^2+b^2c^2\ge2acbd$ but I am not sure how to show that

REPLY [10 votes]: By Lagrange's identity
$$(a^2+b^2)(c^2+d^2)=(ac+bd)^2\color{red}{+(ad-bc)^2}.$$

REPLY [5 votes]: 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 $a^2d^2 + b^2c^2 \geq 2abcd$.

REPLY [4 votes]: Diophantus has already shown that 
$$(a^{2}+b^{2})(c^{2}+d^{2})=(ac+bd)^{2}+(ad-bc)^{2}.$$ This proves the inequality.