Consider some operator $T$ such that $\langle Tv,v \rangle \geq \langle v,v \rangle$ for all $v$. I want to show that $\langle T^k v,v \rangle \geq \langle v,v \rangle$ for all $v$.
My attempt so far has been to consider that $\sigma(p(T)) = p(\sigma(T))$, taking $p = z^k$. But I don't know how to work with this.
This is true if we are working over the complex numbers. Here are two proofs:
Proof 1. Since $T$ is positive it follows that $T$ is self-adjoint. So we can proceed by induction on $k$. It is clearly true for $k=1$. Suppose it's true for all $k \leq n$, where $n \geq 1$. Then $$\langle T^{n+1}v,v \rangle = \langle T^nv, Tv \rangle = \langle T^{n-1}(Tv),Tv \rangle \geq \langle Tv,Tv \rangle = \|Tv\|^2 \geq \|v\|^2$$ Here the first inequality follows from the inductive hypothesis, and the final inequality follows from the fact that $\|v\|^2 \leq \langle Tv,v\rangle \leq \|Tv\|\|v\|$ and thus $\|v\|\leq \|Tv\|$.
Proof 2. For a (perhaps more enlightening) proof, use the (general) spectral theorem to say that $T$ is equivalent to multiplication by some function $f$ on $L^2(X,\mu)$, for some choice of $(X,\mu)$. The given condition on $T$ means precisely that $f\geq 1$ almost everywhere. Then $f^k\geq 1$ almost everywhere, hence $T^k$ will satisfy the above property.
Edit: However this statement becomes false if we work over the real numbers. Indeed, let $T: \Bbb R^2 \to \Bbb R^2$ be defined by $T(x,y)=(2x-2y,2y)$. Then $$\langle T(x,y),(x,y) \rangle = 2x^2 - 2xy + 2y^2 = x^2+y^2 +(x-y)^2 \geq x^2+y^2$$ However, we see that $T^2(x,y) = (4x-8y, 4y)$ so that $\langle T^2(1,1), (1,1) \rangle =0$. Thanks to TZakrevskiy for pointing out this detail.