I am working on the proof of the Kodaira embedding theorem and on one part of this proof you have "$−K_{X}+L^{\otimes k}$ is positive". Since it is part of the requirements, it is clear that $L$ is positive. But generally, is the following statement true?
Let $X$ be a compact Kähler manifold. Then the anticanonical bundle $−K_{X}$ (or $K_{X}^{*}$) is positive.
A holomorphic line bundle $L$ on a compact complex manifold $X$ is positive if $c_1(L) \in H^2(X; \mathbb{Z})$ can be represented by a closed positive $(1, 1)$-form. It follows that $X$ admits a Kähler metric with Kähler form $\omega$ such that $c_1(L) = [\omega]$. In particular, if your claim were true, a compact complex manifold would be Kähler if and only if had positive anticanonical bundle - as we will see, this is far from true.
On a compact Riemann surface $X$, a holomorphic line bundle $L$ has a degree: if $c_1(L) = [\alpha]$, then
$$\deg L = \langle c_1(L), [X]\rangle = \int_X\alpha.$$
In particular, if $L$ is positive, $c_1(L) = [\omega]$ so
$$\deg L = \int_X\omega > 0.$$
On a Riemann surface, $K_X^* = TX$ so
$$\deg K_X^* = \deg TX = \langle c_1(TX), [X]\rangle = \langle e(TX), [X]\rangle = \chi(X)$$
where $e(TX)$ denotes the Euler class of $TX$. We see that if $X$ is a compact Riemann surface with positive anticanonical bundle,
$$\chi(X) > 0.$$
As $\chi(X) = 2 - 2g$ where $g$ is the genus of $X$, the only compact Riemann surface with positive anticanonical bundle is $\mathbb{CP}^1$.
A compact complex manifold with positive anticanonical bundle is called Fano. By the Kodaira embedding theorem, such a manifold is projective, so any compact Kähler, non-projective manifold will also provide a counterexample to your claim. Furthermore, if $K_X^*$ is positive, $K_X$ is negative so $X$ has Kodaira dimension $-\infty$. This provides another class of counterexamples: Kähler manifolds with non-negative Kodaira dimension.