If $A,B$ are positive, commuting operators, then $AB$ is positive. This is because the unique positive $\sqrt{A}$ must also commute with $B$ and, hence,
$$
\langle ABx,x\rangle = \langle \sqrt{A}Bx,\sqrt{A}x\rangle =
\langle B\sqrt{A}x,\sqrt{A}x\rangle \ge 0.
$$
This result is useful in what follows.
Suppose $A$ is selfadjoint. Let $P=\frac{1}{2}(|A|+A)$ and $N=\frac{1}{2}(|A|-A)$, where $|A|$ is the unique positive square root of $A^2$. Then $PN=NP=0$ and $A=P-N$. This is the desired decomposition of $A$, and the trick is to show that $P,N$ are positive operators.
Let $E$ be the orthogonal projection onto $\mathcal{N}(|A|+A)$. Then $(|A|+A)E=0$ gives $E(|A|+A)=0$ by taking adjoints. And $(|A|+A)(|A|-A)=0$ gives $E(|A|-A)=|A|-A$. Hence,
$$
2EA=E(|A|+A)-E(|A|-A) = A-|A| \\
|A| = (I-2E)A \\
2E|A| = E(|A|+A)+E(|A|-A)=|A|-A \\
A = (I-2E)|A|.
$$
These two equations are consistent because $(I-2E)^2=I-4E+4E=I$ establishes $I-2E$ as its own inverse. Taking adjoints of the above equations shows that $E$ commutes with $A$ and with $|A|$, which is useful in what follows. Now the operators $P$ and $N$ may be written as
$$
P=\frac{1}{2}(|A|+A)=\frac{1}{2}(|A|+(I-2E)|A|)=(I-E)|A|, \\
N=\frac{1}{2}(|A|-A)=\frac{1}{2}(|A|-(I-2E)|A|)=E|A|
$$
Because $E$ commutes with $A$, then $E$ must also commute with $A^2$ and, hence, also with $|A|=(A^2)^{1/2}$. By the result of the first paragraph, $P=(I-E)|A|$ and $N=E|A|$ are positive.
Best Answer
Since $T$ is self-adjoint, we have $$ \langle T^{2n+1}\xi,\xi\rangle=\langle T(T^n\xi),T^n\xi\rangle, $$ and since $T$ is positive, this expression is positive.