Proof of Cauchy-Schwarz Inequality for a Positive Semidefinite Quadratic Form on $\mathbb{R}^n$

Write $[n]$ for the initial segment determined by $n$. By Symmetric, Positive Semidefinite, and Positive Definite Real Matrices and Transpose of a Real Matrix, symmetry of $B$ means $B_{ij}=B_{ji}$ for all $i,j\in[n]$, and positive semidefiniteness means $0\le z\cdot(Bz)$ for every $z\in\mathbb{R}^n$. Record that squares are nonnegative in an ordered field: $0\,0=0$ by Zero Products and Elementary Identities in a Field; if $0<t$ then $0<t\,t$ by claim 5 of Elementary Order Arithmetic in an Ordered Field; and if $t<0$ then $0<-t$ by claim 4 of that lemma, so $0<(-t)(-t)=t\,t$.

Claim 1. By claim 4 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum,

$$y\cdot(Bx)=\sum_{i=1}^{n}\sum_{j=1}^{n}B_{ij}\,y_{i}\,x_{j},\qquad x\cdot(By)=\sum_{i=1}^{n}\sum_{j=1}^{n}B_{ij}\,x_{i}\,y_{j},$$

the sums being finite sums in $\mathbb{R}$. Interchanging the order of summation in the first double sum by Interchange of a Finite Double Sum and then renaming the two summation indices turns it into $\sum_{i=1}^{n}\sum_{j=1}^{n}B_{ji}\,y_{j}\,x_{i}$, which equals the second double sum because $B_{ji}=B_{ij}$ and multiplication in $\mathbb{R}$ is commutative.

Claim 2. Put $\alpha=x\cdot(Bx)$, $\beta=x\cdot(By)$ and $\gamma=y\cdot(By)$, so that $0\le\alpha$ and $0\le\gamma$. Let $t\in\mathbb{R}$. By claim 3 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum and claims 2, 4 and 5 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$, together with claim 1 above,

$$0\le(x+t\,y)\cdot\bigl(B(x+t\,y)\bigr)=\alpha+t\,\beta+t\,\beta+t^{2}\gamma. \tag{$*$}$$

Since $\le$ is total, either $\gamma=0$ or $0<\gamma$.

Suppose $\gamma=0$, so that $t^{2}\gamma=0$ by Zero Products and Elementary Identities in a Field and $(*)$ reads $0\le\alpha+t\,\beta+t\,\beta$ for every $t\in\mathbb{R}$. Assume for contradiction that $\beta\ne0$. Setting $2=1+1$, we have $0<2$ by claim 8 of Elementary Order Arithmetic in an Ordered Field, so $2\ne0$, and therefore $2\,\beta\ne0$ by Zero Products and Elementary Identities in a Field. Taking $t=-(\alpha+1)(2\,\beta)^{-1}$ in $(*)$, and using $t\,\beta+t\,\beta=t\,(2\beta)$, gives $0\le\alpha-(\alpha+1)=-1$, which contradicts $0<1$ (claim 6 of Elementary Order Arithmetic in an Ordered Field) by claim 4 of that lemma. Hence $\beta=0$, so $\beta^{2}=0$ and $\alpha\gamma=0$ by Zero Products and Elementary Identities in a Field, and the asserted inequality reads $0\le0$.

Suppose now $0<\gamma$. By claim 7 of Elementary Order Arithmetic in an Ordered Field, $\gamma^{-1}$ exists and $0<\gamma^{-1}$. Taking $t=-\beta\,\gamma^{-1}$ in $(*)$ and simplifying with the field axioms,

$$t\,\beta+t\,\beta+t^{2}\gamma=-\beta^{2}\gamma^{-1}-\beta^{2}\gamma^{-1}+\beta^{2}\gamma^{-1}=-\beta^{2}\gamma^{-1},$$

so $0\le\alpha-\beta^{2}\gamma^{-1}$, that is $\beta^{2}\gamma^{-1}\le\alpha$ by claim 3 of Elementary Arithmetic in an Ordered Field. Multiplying by the nonnegative element $\gamma$ using claim 5 of Elementary Arithmetic in an Ordered Field, and using $\gamma\,(\beta^{2}\gamma^{-1})=\beta^{2}$, gives $\beta^{2}\le\gamma\,\alpha=\alpha\gamma$.

Claim 3. Suppose $x\cdot(Bx)=0$ and let $i\in[n]$. Let $e_{i}\in\mathbb{R}^n$ be the point whose $i$th coordinate is $1$ and whose remaining coordinates are $0$. Applying claim 2 with $y=e_{i}$, and using $0\,s=0$ from Zero Products and Elementary Identities in a Field,

$$\bigl(x\cdot(Be_{i})\bigr)^{2}\le 0\cdot\bigl(e_{i}\cdot(Be_{i})\bigr)=0 .$$

Since squares are nonnegative, $\bigl(x\cdot(Be_{i})\bigr)^{2}=0$, and hence $x\cdot(Be_{i})=0$ by Zero Products and Elementary Identities in a Field. By claim 1, $e_{i}\cdot(Bx)=x\cdot(Be_{i})=0$.

Finally, by Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$, $e_{i}\cdot(Bx)=\sum_{k=1}^{n}(e_{i})_{k}(Bx)_{k}$; every summand with $k\ne i$ vanishes, because $(e_{i})_{k}=0$ and $0\,s=0$, so claim 7 of Properties of Finite Sums evaluates this sum as $(e_{i})_{i}(Bx)_{i}=(Bx)_{i}$. Hence $(Bx)_{i}=0$ for every $i\in[n]$, that is $Bx=0_{\mathbb{R}^n}$ by The Origin of $\mathbb{R}^n$.