# Is this proof of Cauchy Schwarz inequality circular or valid?

I'm a college freshman learning linear algebra on my own, and I'm in the section on inner products. I noticed a proof of the Cauchy Schwarz inequality for vectors in my book, and it seems to contain non-intuitive things about it by using a quadratic function - the book even says a clever, non-intuitive trick is required. I thought there might be a straightforward proof of the theorem, so I tried to construct a straightforward proof of it on my own, but I wasn't sure if this proof was correct. This is my proof below.

Theorem: If $u$ and $v$ are vectors in an inner product vector space $V$, then

$$\vec u\cdot\vec v\le\|\vec u\|\|\vec v\|.\tag1$$

Proof: Let $V$ be an inner product vector space. Since $(\vec u\cdot\vec u)^{1/2} = \|\vec u\|$ by definition, $V$ is also a vector space equipped with a norm.

By one of the axioms for vector norms for a normed vector space,

$$\|\vec u+\vec v\|\le\|\vec u\| + \|\vec v\|.\tag2$$

We can square both sides of (2) to get

$$\|\vec u+\vec v\|^2 \le (\|\vec u\| + \|\vec v\|)^2\tag3$$

Since we can express the norm of $\vec u$ in terms of inner products, we can square both sides of (3) and expand the left hand side in inner products and the right hand side as a binomial expansion.

\begin{align} \|\vec u+\vec v\|^2 &= (\vec u+\vec v)\cdot(\vec u+\vec v) = \vec u\cdot(\vec u+\vec v) + \vec v\cdot(\vec u+\vec v) =\\ &= \|\vec u\|^2 + \vec u\cdot\vec v + \vec u\cdot\vec v + \|\vec v\|^2 =\|\vec u\|^2 + \|\vec v\|^2 + 2[\vec u\cdot\vec v]. \end{align} The right hand side expands as $$(\|\vec u\| + \|\vec v\|)(\|\vec u\| + \|\vec v\|) = \|\vec u\|^2 + \|\vec v\|^2 + 2\|\vec u\|\|\vec v\|.$$

We can now subtract like terms from both sides of equation (3), and obtain that

$$2[\vec u\cdot\vec v] \le 2\|\vec u\|\|\vec v\|,\quad\text{or}\quad \vec u\cdot\vec v \le \|\vec u\|\|\vec v\|.\qquad \text{QED}$$

Is this proof circular, or is it valid? Thanks.

• The triangle inequality is a consequence of Cauchy Schwarz inequality. – user260717 Aug 16 '15 at 21:11

It's circular.

Since $(\vec u\cdot\vec u)^{1/2} = \|\vec u\|$ by definition, $V$ is also a vector space equipped with a norm.

$(\vec u\cdot\vec u)^{1/2} = \|\vec u\|$ defines what the symbol $\|\vec u\|$ means. This symbol is commonly used for a norm, but the definition alone doesn't make it one. You have to prove that it satisfies the definition of the word "norm". You haven't done that, so your proof is certainly incomplete. And proving that $\|\vec u\|$, as defined above, satisfies the triangle inequality, is usually done by using Cauchy-Schwarz. Hence the circularity.

• Unfortunately the en.wikipedia.org/wiki/Duck_test doesn't count as a valid mathematical proof. It looks like a norm, but you haven't shown that it is a norm. – alephzero Aug 16 '15 at 23:24

Your algebra shows the equivalence between

• Cauchy-Schwarz inequality
• the triangle inequality for the distance defined as $|XY| = \sqrt{(X-Y,X-Y)}$

If either inequality holds, so does the other.

Proving that at least one of them does hold is not addressed by the calculation in the question.