Proving Schwarz inequlity.

Question

Is the following proof correct?

The Schwarz inequality states: $$ x_1 \cdot y_1 + x_2 \cdot y_2 \leq \sqrt{(x_1^2+x_2^2)\cdot(y_1^2+y_2^2)} $$ From $$ (x_1^2+x_2^2)\cdot(y_1^2+y_2^2) $$ We can deduce: $$ (x_1^2+x_2^2)\cdot(y_1^2+y_2^2) = (x_1y_1+x_2y_2)^2 + (x_1y_2-x_2y_1)^2 $$ We can rewrite it in form: $$ (x_1y_2-x_2y_1) = \sqrt{(x_1^2+x_2^2)\cdot(y_1^2+y_2^2) - (x_1y_1+x_2y_2)^2} $$ Meaning that $$ (x_1^2+x_2^2)\cdot(y_1^2+y_2^2) - (x_1y_1+x_2y_2)^2 \geq 0 $$ From which the Schwarz inequality can be easily deduced. $$ \sqrt{(x_1^2+x_2^2)\cdot(y_1^2+y_2^2)} \geq x_1y_1+x_2y_2 $$

Answer 1

Yes it is correct even if from here

$$(x_1^2+x_2^2)\cdot(y_1^2+y_2^2) = (x_1y_1+x_2y_2)^2 + (x_1y_2-x_2y_1)^2$$

we can deduce directly that

$$(x_1^2+x_2^2)\cdot(y_1^2+y_2^2) - (x_1y_1+x_2y_2)^2=(x_1y_2-x_2y_1)^2 \geq 0$$

This proof is nice but it is limited to n=2, for a general proof we can start from

$$\sum x_iy_i\le \sqrt{\sum x_i^2\sum y_i^2}$$

and observe that WLOG for homogenity we can rescale $x$ and $y$ such that $\sum x_i^2=\sum y_i^2=1$ and thus we need to prove that

$$\sum x_iy_i\le 1$$

then observe that

$x_iy_i\le \frac12(x_i^2+y_i^2)\iff (x_i-y_i)^2\ge 0$

then

$$\sum x_iy_i\le \frac12 \sum (x_i^2+y_i^2) =\frac12 (\sum x_i^2+\sum y_i^2)=1$$

Answer 2

Yes, the proof is correct! But the line $$ (x_1y_2-x_2y_1) = \sqrt{(x_1^2+x_2^2)\cdot(y_1^2+y_2^2) - (x_1y_1+x_2y_2)^2} $$ is redundant; you can shorten your proof by skipping it.