Suppose $⟨Tv, v⟩ ≥ 0$ for all $v$, with equality iff $v=0$. If $Tv = 0$, then $v=0$, so $\ker T = \{0\}$, which means $T$ is injective and therefore invertible as well.
If $T$ is positive and invertible, by the spectral theorem, $V$ has an orthonormal basis that consists of eigenvectors $u_i$ of $T$, with nonnegative eigenvalues $λ_i$. We can write $v = \sum ⟨v, u_i⟩ u_i$ and $Tv = \sum ⟨v, u_i⟩ λ_i u_i$. Therefore, $⟨Tv, v⟩ = \sum λ_i \left|⟨v, u_i⟩\right|^2$ which is 0 iff $⟨v, u_i⟩ = 0$ for all $i$, and $v=0$. ($T$ is positive so we already have $⟨Tv, v⟩ ≥ 0$ by definition.)