Proof: Matrix has just non-negative real eigenvalues

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$\textbf{So my given problem is:}$

$Let\,\, A \in \mathbb{C}^{n \,\times \,n}\,\, be\,\, such\,\, that\,\,\\ \forall x\in \mathbb{C}^n : \langle\,Ax,x\rangle \geq 0 \\ where \,\, \langle\,\cdot,\cdot\rangle \,\, is \,\, the \,\, standard \,\, inner \,\, product \,\, of \,\, \mathbb{C}^n.\,\, Show \,\, that\,\,\\ spec(A) \subset [0, \infty).$

$\textbf{My proof:}$

We have:

$\langle\,Ax,x\rangle \geq 0$

Now let $\lambda$ be e-value of A and x the corresponding e-vector then the following equality holds.

$Ax = \lambda x$

Then we obtain:

$\langle\,\lambda x,x\rangle \geq 0$

$\Leftrightarrow \lambda \langle\,x,x\rangle \geq 0$

Since $\langle\,x,x\rangle$ is always a non-negativ real number $\forall x \in \mathbb{C}^n$. So the whole term then is greater or equal to zero if and only if $\lambda \geq 0$.

$\Rightarrow spec(A) \subset [0,\infty). \square$

$\textbf{NOTE:}$

My professor defined $\langle\,\lambda x,x\rangle = \lambda \langle\,x,x\rangle$

Now I'm looking for verification of that proof.