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:
- $|\boldsymbol{u} \cdot \boldsymbol{v}| \leq\|\boldsymbol{u}\|\|\boldsymbol{v}\|$, 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
$$
\|\boldsymbol{v}\|^2\|\boldsymbol{u}\|^2 \geq(\boldsymbol{u} \cdot \boldsymbol{v})^2, \Longrightarrow|\boldsymbol{u} \cdot \boldsymbol{v}| \leq\|\boldsymbol{u}\|\|\boldsymbol{v}\| .
$$
Best Answer
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.