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

Question

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

Answer 1

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

Answer 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 $a^2d^2 + b^2c^2 \geq 2abcd$.

Answer 3

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