Question on positive definiteness of non homogeneous quadratic form

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I'm having trouble understanding a proposition from a semidefinite-programming textbook. It goes as follows:

Let $Q$ be a quadratic function of $x \in \mathbb{R}^n$ given by

$$Q(x) = x^TAx+2b^Tx+c$$

which can be written as

$$Q(x)= \begin{bmatrix} x\\1 \end{bmatrix}^T \begin{bmatrix} A & b\\b^T & c \end{bmatrix} \begin{bmatrix} x\\1 \end{bmatrix}$$

where $A=A^T\in\mathbb{R}^{n\times n}$, $b\in \mathbb{R}^n$ and $c\in\mathbb{R}$. Then $Q(x)\geq0, \forall x\in\mathbb{R}^n$ if and only if

$$ \begin{bmatrix} A & b\\b^T & c \end{bmatrix} \succeq0 $$

The if part is pretty straightforward, but I'm having trouble with the only if.

My approach was to consider the vector $\tilde{x}\in\mathbb{R}^{n+1}$, to write

$$\tilde{x}^T \begin{bmatrix} A & b\\b^T & c \end{bmatrix} \tilde{x} \geq 0$$

If we divide this equation by $\tilde{x}_4^2$ (the fourth element of the vector $\tilde{x}$) and call $x = \frac{1}{\tilde{x}_4}(\tilde{x}_1, \tilde{x}_2, \tilde{x}_3)$, we find the same equation as above, but this means that the implication is only valid for $\tilde{x}_4 \neq 0$.

The proposition goes even further and says that if the inequality is strict, the second strict inequality is only sufficient for the positivity of $Q(x)$, and no longer necessary.

Could anyone enlighten me on this matter, please? Thanks in advance.