Assume $T=T^{\star}\in\mathcal{L}(H)$, where $H$ is a complex Hilbert Space.
Implication 1: Show $(Tx,x) \ge 0$ for all $x \in H$ implies $\sigma(T)\subseteq [0,\infty)$.
To do this, assume that $(Tx,x) \ge 0$ for all $x \in H$, and let $\lambda < 0$. Then
$$
0 \le -\lambda(x,x) \le ((T-\lambda I)x,x)
$$
implies
$$
|\lambda|\|x\|^{2} \le \|(T-\lambda I)x\|\|x\|,\\
|\lambda|\|x\| \le \|(T-\lambda I)x\|.
$$
Therefore $T-\lambda I$ is injective for $\lambda < 0$, and its range is closed because the inverse is bounded. But the range is dense because
$\mathcal{R}(T-\lambda I)^{\perp}=\mathcal{N}(T-\lambda I)=\{0\}$. Therefore $\lambda \in \rho(A)$ for $\lambda < 0$.
Implication 2: Show $\sigma(T)\subseteq [0,\infty)$ implies $(Tx,x) \ge 0$ for all $x \in H$.
There are many techniques for proving this implication, but I'll assume you know that the norm $\|T\|$ is the same as the spectral radius $r_{\sigma}(T)$ for a bounded selfadjoint operator $T$. Therefore, $\sigma(T)\subseteq [0,\|T\|]$. So, $\sigma(T-\|T\|/2)\subseteq [-\|T\|/2,\|T\|/2]$ which then implies that
$$
\left\|T-\frac{\|T\|}{2}I\right\| \le \frac{\|T\|}{2}.
$$
Consequently,
$$
\left|\left(\left(T-\frac{\|T\|}{2}I\right)x,x\right)\right| \le \frac{\|T\|}{2}\|x\|^{2},\\
%% \left|(Tx,x)-\frac{\|T\|}{2}\|x\|^{2}\right| \le \frac{\|T\|}{2}\|x\|^{2},\\
%% -\frac{\|T\|}{2}\|x\|^{2} \le (Tx,x) -\frac{\|T\|}{2}\|x\|^{2}
%% \le \frac{\|T\|}{2}\|x\|^{2},\\
\implies 0 \le (Tx,x) \le \|T\|\|x\|^{2}.
$$