Equivalence of ample divisor

Question

Let $X$ be a smooth complex projective variety. Having ample anticanonical class is equivalent to having $-K_{X} > 0$? How to prove that? Does it hold for any ample divisor in $X$?

Answer 1

Yes, the equivalence holds for every ample divisor on $X$.
More precisely, if $X$ is a complex compact Kähler manifold and if $L$ is a holomorphic line bundle on $X$ then we have the following celebrated equivalence due to Kodaira : $$L\; \text {is ample} \iff L \; \text {is positive } $$ If such an $L$ exists, then $X$ is of course automatically a smooth projective algebraic variety.

Let me remind you that:
1) $L$ is said to be ample if there exists a positive integer $m$ such that $M=L^{\otimes m}$ is very ample, which in turn means that there exists for some positive $N$ a closed embedding $i: X\hookrightarrow \mathbb P^N_\mathbb C$ such that $$M\cong i^*(\mathcal O_{\mathbb P^N} (1))$$ 2) $L$ is said to be positive if it can be endowed with a hermitian metric whose associated $(1,1)$ form $\omega$ is closed [meaning $d \omega=0$] and positive [meaning $\omega(v,iv)\gt 0$ for all $x\in X$ and all nonzero tangent vectors $0\neq v\in T_x(X)$].

Your question is the special case where $L$ is the anti-canonical line bundle $L=\omega_X^{-1}=\mathcal O_X(-K_X)$.