That is true. Basically it is a consequence of the following fact:
Let $D$ be a nef and big divisor on an irreducible projective variety $X$. Then there exist an effective divisor $E$ and a rational number $0 <\epsilon\ll 1$ such that $D-\epsilon E$ is ample.
Proof: Let $D$ be a nef and big divisor. Since $D$ is big, by \cite[Corollary 2.2.6]{La}, there exist an ample divisor $A$, an effective divisor $E$, and a positive integer $k$ such that $kD\equiv A+E$. If $h>k$ we can write $hD\equiv (h-k)D+A+E$. The divisor $D^{'} = (h-k)D+A$ is a sum of a nef and an ample divisor. Therefore $D^{'}$ is ample. If $\epsilon = \frac{1}{h}$ we get that
$$D-\epsilon E\equiv \epsilon D^{'}$$
is ample.
Now, it is enough to apply this to $D = -K_X$. What you get in the end is the follwing:
Let $X$ be an irreducible, projective variety with at most klt singularities. If $X$ is weak Fano then $X$ is log Fano.