Schwarz inequality is Cauchy–Schwarz inequality?

[rfrederic]

I found many information showed Schwarz inequality and Cauchy–Schwarz inequality are same on books and internet, but my teacher's material shows that:
Schwarz inequality:
$\left\|[x,y]\right\|\leq\left\|x\right\|+\left\|y\right\|$

Cauchy–Schwarz inequality:
$\left\|[x,y]\right\|\leq\left\|x\right\|\left\|y\right\|$


They seem to be different on material, and I had sent email to teacher but having no reply.
Therefore my question is "Are Schwarz inequality and Cauchy–Schwarz inequality same?"

 

[mfb]

Those do not look like inequalities to me. And the first one looks wrong, independent of the inequality sign.
Maybe you mean $||x+y|| \leq ||x||+||y||$, but that is the triangle inequality. It follows from the Cauchy–Schwarz inequality if the norm is induced by a scalar product.

 

[rfrederic]

Those do not look like inequalities to me. And the first one looks wrong, independent of the inequality sign.
Maybe you mean $||x+y|| \leq ||x||+||y||$, but that is the triangle inequality. It follows from the Cauchy–Schwarz inequality if the norm is induced by a scalar product.

Sorry about used wrong symbol, and I have modified.
I know triangle inequality.

But the question still is "are Schwarz inequality and Cauchy–Schwarz inequality same?"


Thanks for your reply.

 

[micromass]

Your Schwarz inequality simply seems false. In $\mathbb{R}$, we have $[x,y]=xy$. But it is certainly not the case that

$$|2\cdot 3|\leq |2|+|3|$$

 

[Vargo]

It looks like a typo to me. The books and the internet are right I think.

 

[rfrederic]

Your Schwarz inequality simply seems false. In $\mathbb{R}$, we have $[x,y]=xy$. But it is certainly not the case that

$$|2\cdot 3|\leq |2|+|3|$$

Well, do you mean that:
Schwarz inequality
= Cauchy–Schwarz inequality
= $\left\|[x,y]\right\|\leq\left\|x\right\|\left\|y\right\|$?

It looks like a typo to me. The books and the internet are right I think.

I think so either, thus I want to figure it out.


Both of your answers are helpful, thanks.