# Proving the Cauchy-Schwarz inequality, spivak's calculus

I'm working on a problem concerning the Cauchy-Schwarz inequality from Spivak's calculus 3rd edition. The problem consists of completing three equivalent proofs of the inequality. In one of the proofs the inequality can be deduced from the fact $(x_1^2+x_2^2)(y_1^2+y_2^2)=(x_1y_1+x_2y_2)^2+(x_1y_2-x_2y_1)^2$ for real numbers $x_1,x_2,y_1$ and $y_2$. Manipulation of this equation together with facts about inequalities leads one to $|x_1y_1+x_2y_2|\leq \sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2}$ . The other proof uses $2xy\leq x^2+y^2$ and the substitutions $x:=\frac{x_A}{\sqrt{x_1^2+x_2^2}}$ and $y:=\frac{y_A}{\sqrt{y_1^2+y_2^2}}$, first for $A=1$ and then for $A=2$. Here's where my difficulties arise: when doing the second proof, I can arrive at the inequality, but not for the absolute value of $x_1y_1+x_2y_2$. Can anyone work the second proof and see how the absolute value comes into play?

Thanks

• Can you show how you obtain second proof? – Idonknow Jan 5 '18 at 3:06

## 1 Answer

The same proof procedure can also be applied for $-2xy\le x^2+y^2$

You will find $$x_1y_1+x_2y_2 \leq \sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2}$$ and $$-(x_1y_1+x_2y_2)\leq \sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2}$$

so $$|x_1y_1+x_2y_2|\leq \sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2}$$

• Oh, you're right! I forgot that inequality :P Thank you for your answer! – Owl Tree Jan 5 '18 at 4:06