I'm quite new to math proofs,I can't understand why the part where we set $\alpha=\|\boldsymbol{v}\|^2$ e $\beta=-\boldsymbol{u} \cdot \boldsymbol{v}$ works , why are proofs like those possible and valid? By setting alpha and beta to these values aren't we proving the theorem just for the case where $\alpha=\|\boldsymbol{v}\|^2$ e $\beta=-\boldsymbol{u} \cdot \boldsymbol{v}$, the proof isn't generalizing enough or am I missing something?

Theorem : If $(V, \cdot)$ is an euclidean vector space (real), then $\forall \boldsymbol{u}, \boldsymbol{v} \in V$, we have:

  1. $|\boldsymbol{u} \cdot \boldsymbol{v}| \leq\|\boldsymbol{u}\|\|\boldsymbol{v}\|$, Cauchy-Schwarz inequality.

Proof. Let us first prove the Cauchy-Schwarz inequality. It is clear that the inequality is verified if at least one of the two vectors is null. We therefore assume that they are both nonzero. Let us consider $\boldsymbol{w}=\alpha \boldsymbol{u}+\beta \boldsymbol{v}, \operatorname{with} \alpha, \beta \in \mathbb{R}$, $$ \boldsymbol{w} \cdot \boldsymbol{w}=(\alpha \boldsymbol{u}+\beta \boldsymbol{v}) \cdot(\alpha \boldsymbol{u}+\beta \boldsymbol{v})=\alpha^2\|\boldsymbol{u}\|^2+\beta^2\|\boldsymbol{v}\|^2+2 \alpha \beta \boldsymbol{u} \cdot \boldsymbol{v} \geq 0 . $$

*** then taking $\alpha=\|\boldsymbol{v}\|^2$ e $\beta=-\boldsymbol{u} \cdot \boldsymbol{v}$, we get $$ \|v\|^4\|\boldsymbol{u}\|^2+(\boldsymbol{u} \cdot \boldsymbol{v})^2\|v\|^2-2\|v\|^2(\boldsymbol{u} \cdot \boldsymbol{v})^2=\|\boldsymbol{v}\|^2\left(\|\boldsymbol{v}\|^2\|\boldsymbol{u}\|^2-(\boldsymbol{u} \cdot \boldsymbol{v})^2\right) \geq 0 . $$

Since $v \neq 0$, we can divide by $\|v\|^2$, and get the inequality $$ \|\boldsymbol{v}\|^2\|\boldsymbol{u}\|^2 \geq(\boldsymbol{u} \cdot \boldsymbol{v})^2, \Longrightarrow|\boldsymbol{u} \cdot \boldsymbol{v}| \leq\|\boldsymbol{u}\|\|\boldsymbol{v}\| . $$


3 Answers 3


In the proof provided, we see that the author have shown the truthness of $$\alpha^2\|\boldsymbol{u}\|^2+\beta^2\|\boldsymbol{v}\|^2+2 \alpha \beta \boldsymbol{u} \cdot \boldsymbol{v} \geq 0 $$ for all $\boldsymbol{u},\boldsymbol{v}\in V$ and for all $\alpha,\beta\in\mathbb{R}$.

In the logical form, we can write $$\left(\forall\boldsymbol{u},\boldsymbol{v}\in V\right)\left(\forall\alpha,\beta\in\mathbb{R}\right)\left(\alpha^2\|\boldsymbol{u}\|^2+\beta^2\|\boldsymbol{v}\|^2+2 \alpha \beta \boldsymbol{u} \cdot \boldsymbol{v} \geq 0\right).$$

We know that if $\forall xP(x)$ is true, then $P(x_0)$ is true, because the truthness of $\forall xP(x)$ means that $P(x)$ is true no matter what value is assigned to $x$. Therefore, we can choose any value, say $x_0$, for $x$ and conclude that $P(x_0)$ is true. Logicians call this rule of inference universal instantiation.

Thus, if $$\left(\forall\boldsymbol{u},\boldsymbol{v}\in V\right)\left(\forall\alpha,\beta\in\mathbb{R}\right)\left(\alpha^2\|\boldsymbol{u}\|^2+\beta^2\|\boldsymbol{v}\|^2+2 \alpha \beta \boldsymbol{u} \cdot \boldsymbol{v} \geq 0\right)$$ is true, then we can choose $\alpha_0=\|\boldsymbol{v}\|^2\in\mathbb{R}$ and $\beta_0=-\boldsymbol{u} \cdot \boldsymbol{v}\in\mathbb{R}$ to conclude that $$\left(\forall\boldsymbol{u},\boldsymbol{v}\in V\right)\left(\alpha_0^2\|\boldsymbol{u}\|^2+\beta_0^2\|\boldsymbol{v}\|^2+2 \alpha_0 \beta_0 \boldsymbol{u} \cdot \boldsymbol{v} \geq 0\right)$$ is also true.


The proof's explanation

The inequality

$$ \left\|\mathbf{v}\right\|^{2}\beta^{2} + 2\alpha\left(\mathbf{u}\cdot\mathbf{v}\right)\beta + \alpha^{2}\left\|\mathbf{u}\right\|^{2} \geq 0 $$

represent a region bounded by a quadratic function in $\beta$ (or $\alpha$). Since the coefficient of $\beta^{2}$ is positive, the quadratic function has a global minima at

$$ \beta = \frac{-\alpha\left(\mathbf{u}\cdot\mathbf{v}\right)}{\left\|\mathbf{v}\right\|^{2}} \iff \frac{\beta}{\alpha}=\frac{-\mathbf{u}\cdot\mathbf{v}}{\left\|\mathbf{v}\right\|^{2}} $$

and that's why the author chose $\alpha=\left\|\mathbf{v}\right\|^{2}$ and $\beta=-\mathbf{u}\cdot\mathbf{v}$

Alternative choice for $\alpha$ and $\beta$

As mentioned before, the inequality can also represent a region bounded by a quadratic function in $\alpha$. Similarly, the coefficient of $\alpha^{2}$ is positive, so the quadratic function has a global minima at

$$ \alpha=\frac{-\beta\left(\mathbf{u}\cdot\mathbf{v}\right)}{\left\|\mathbf{u}\right\|^{2}} \iff \frac{\alpha}{\beta} = \frac{-\mathbf{u}\cdot\mathbf{v}}{\left\|\mathbf{u}\right\|^{2}} $$

So you can choose $\alpha = -\mathbf{u}\cdot\mathbf{v}$ and $\beta=\left\|\mathbf{u}\right\|^{2}$, and will still prove the CS inequality.

My preferred proof

I dislike the 'arbitrary' choice of $\alpha$ and $\beta$ too. So this is my preferred proof. The following holds:

$$ \left\|x\mathbf{u}+\mathbf{v}\right\|^{2} \geq 0 $$

From which we have

$$ \left\|\mathbf{u}\right\|^{2}x^{2} + 2\left(\mathbf{u}\cdot\mathbf{v}\right)x + \left\|\mathbf{v}\right\|^{2} \geq 0 $$

Due to the inequality, the discriminant of the quadratic function on the LHS is non-positive:

$$ 4\left(\mathbf{u}\cdot\mathbf{v}\right)^{2}-4\left\|\mathbf{u}\right\|^{2}\left\|\mathbf{v}\right\|^{2}\leq 0 $$

which implies the CS inequality

$$ \mathbf{u}\cdot\mathbf{v}\leq \left\|\mathbf{u}\right\|\left\|\mathbf{v}\right\| $$


In the proof, the vector $w$ depends on the numbers $\alpha$ and $\beta$. The point is that no matter how we choose $\alpha$ and $\beta$, the critical inequality $w\cdot w \ge 0$ holds.

I'll admit the proof is unsatisfying in the sense that those particular choices of $\alpha$ and $\beta$ come out of thin air, but once we've found them, the rest of the proof is just algebraic manipulation of $w\cdot w\ge 0$ until we arrive at the desired endpoint that $|u\cdot v|\le |u|\cdot|v|$.

  • $\begingroup$ Thank you for answering, so in any math proof picking values like that will be valid? If i understood correctly, we chose this w⋅w≥0 and then chose some values so that the desired inequality comes out, but to me this feels a bit like cheating Sorry if I ask those question but it's literally the first time I approach proofs $\endgroup$
    – Leit22
    Nov 28, 2023 at 14:37
  • 2
    $\begingroup$ @Leit22 - Note that the result you are proving does not involve any such constants $\alpha, \beta$. The proof first proves an intermediary result that does involve arbitrary constants $\alpha, \beta$. Because they are arbitrary, that result holds for any values of these two constants. This includes the values $\alpha = \|\boldsymbol v\|^2$ and $\beta = -\boldsymbol u\cdot \boldsymbol v$, which then proves the Cauchy-Schwarz inequality. $\endgroup$ Nov 29, 2023 at 3:06

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