= Solution
For a <positive operator>, $\langle Tx,x\rangle\geq0$. We show directly that every nonreal or negative real $\lambda$ lies in the <resolvent set>.
If $\operatorname{Im}\lambda\ne0$, the <Cauchy-Schwarz inequality> and reality of $\langle Tx,x\rangle$ give
$$
\|(T-\lambda I)x\|\|x\|\geq|\langle(T-\lambda I)x,x\rangle|\geq|\operatorname{Im}\lambda|\|x\|^2.
$$
Thus $T-\lambda I$ is bounded below, hence injective with closed range: a convergent sequence of outputs has Cauchy preimages, whose limit maps to that output. Its adjoint is $T-\bar\lambda I$, which has trivial <kernel> by the same estimate. The identity $\operatorname{Ran}(A)^\perp=\ker A^*$, proved by the same inner-product calculation as in (i), makes its range dense. Being closed and dense, it is all of $X$, and the lower bound makes the inverse bounded.
If $\lambda<0$ is real, positivity gives $\langle(T-\lambda I)x,x\rangle\geq(-\lambda)\|x\|^2$, so the same argument applies with lower bound $-\lambda$. Here the operator is <self-adjoint>, so its range is dense because its own <kernel> is zero. Hence
$$
\boxed{\sigma(T)\subseteq[0,\infty).}
$$
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