Proof of Cauchy Schwarz

Question

In real analysis I we saw the following statement that was called the Cauchy - Schwarz inequality:

$|\vec{x}\cdot\vec{y}| \leq \lVert \vec{x} \rVert \lVert \vec{y} \rVert $.

This is the begin of the proof and there is a reasoning I don't understand:

If $\vec{y} = \vec{0}$, than both sides of the inequality are zero. Now, assume $\vec{y} \neq \vec{0}$. For every $t \in \mathbb{R}$ we have:

$0 \leq (\vec{x} + t\vec{y}) \cdot (\vec{x} + t\vec{y}) $. But why is this inequality true, I don't see the right assumptions to know this for sure?

For every vector $\vec v$, $\vec v.\vec v=\left|\vec v\right|^2\geqslant0$. — José Carlos Santos, Oct 14 '23 at 13:44

Badshah Khan · Accepted Answer · 2023-10-14T14:14:54.823

1

This is because of the following definition of norm. i.e for any vector $\vec{x}$, $\vec{x}.\vec{x} = ||\vec{x}||^2$ and square of any real number is always greater or equal to 0.

In your case, $(\vec{x}+t\vec{y})\cdot(\vec{x}+t\vec{y}) = || \vec{x}+t\vec{y} ||^2 \geq 0$

edited Oct 14 '23 at 14:14

answered Oct 14 '23 at 13:55

Badshah Khan

376

1

Actually, it is not because the norm is non-negative, but because $(x+ty)\cdot(x+ty) = |x+ty|^2$ is the square of a real number, which is always non-negative – Didier Oct 14 '23 at 14:01
thanks @Didier. Edited... – Badshah Khan Oct 14 '23 at 14:16

Proof of Cauchy Schwarz

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