Proving $xy+uv\leq\sqrt{x^2+u^2}\sqrt{y^2+v^2}$

Question

How do I prove the Cauchy–Schwarz inequality?

$$xy+uv\leq \sqrt{x^2+u^2} \sqrt{y^2+v^2}$$ with $x,y,u,v \in \mathbb{R} $

Answer 1

If $$xy+uv <0$$ it is true.

if $$xy+uv\ge 0$$ we take the square and prove that

$$2xyuv\le x^2v^2+y^2u^2$$ which is true since $$x^2v^2+y^2u^2-2xyuv=(xv-yu)^2$$

Answer 2

Of course: $$ (x^2+u^2)(y^2+v^2) = (xy+uv)^2\color{red}{+(xv-uy)^2} $$ is Lagrange's identity, straightforward to check.

Answer 3

if $$xy+uv<0$$ then our inequality is true. In the other case we get by squaring $$x^2y^2+u^2v^2+2xyuv\le x^2y^2+y^2u^2+x^2v^2+u^2v^2$$ so we get $$(yu-xv)^2\geq 0$$

Answer 4

Since the inequality is homogeneous for the two set of the letters we may assume that $x^2+u^2=1$ and $y^2+v^2=1.$ Then we have to prove inequality $$ xy+uv \leq 1. $$ We have $2xy \leq x^2+y^2$ and $2uv\leq u^2+v^2.$ Thus $$ 2xy+2uv \leq x^2+y^2+u^2+v^2=2 \implies xy+uv \leq 1. $$