The calculation in local coordinates is not too hard and it works in any dimension $n$, namely if $c_1(M)=0$ then $\int_M c_2(M)\wedge\omega^{n-2}\geq 0$ for any K"ahler metric $\omega$. Of course you are free to assume that the Kahler metric $\omega$ is Ricci-flat (by Yau's theorem), and in this case the integral is equal to the $L^2$ norm of the Riemann curvature tensor up to a factor.

To see this, let $\Omega_i^j=\frac{\sqrt{-1}}{2\pi} R^j_{i k\overline{\ell}}dz^k\wedge d\overline{z}^\ell$ denote the curvature form, then by Chern-Weil theory the form $\frac{1}{4\pi^2}Tr(\Omega\wedge\Omega)=\frac{(\sqrt{-1})^2}{4\pi^2}\sum_{k,i} \Omega_i^k\wedge\Omega_k^i =\frac{(\sqrt{-1})^2}{4\pi^2}R^k_{ip\overline{q}} R^i_{kr\overline{s}}dz^p\wedge d\overline{z}^q\wedge dz^r\wedge d\overline{z}^s$ represents $c_1^2(M)-2c_2(M)=-2c_2(M)$. You can assume that the metric $\omega$ is the identity at one point, and then at that point you compute $$n(n-1)Tr(\Omega\wedge\Omega)\wedge\omega^{n-2}=\sum_{p\neq r}(R^k_{ip\overline{p}} R^i_{kr\overline{r}}-R^k_{ip\overline{r}} R^i_{kr\overline{p}})\omega^n

=\sum_{p,r}(R^k_{ip\overline{p}} R^i_{kr\overline{r}}-R^k_{ip\overline{r}} R^i_{kr\overline{p}})\omega^n

=(|\textrm{Ric}|^2-|\textrm{Rm}|^2)\omega^n=- |\textrm{Rm}|^2\omega^n.$$ Integrating this you get what you want, with equality if and only if $\omega$ is flat and so $M$ is finitely covered by a complex torus.

A reference for a general statement (I think this is not the earliest paper where this result appears) is:

Chen, Bang-yen; Ogiue, Koichi, Some characterizations of complex space forms in terms of Chern classes, Quart. J. Math. Oxford Ser. (2) 26 (1975), no. 104, 459–464.

The calculation in local coordinates is not too hard and it works in any dimension $n$, namely if $c_1(M)=0$ then $\int_M c_2(M)\wedge\omega^{n-2}\geq 0$ for any Kahler metric $\omega$. Of course you are free to assume that the Kahler metric $\omega$ is Ricci-flat (by Yau's theorem), and in this case the integral is equal to the $L^2$ norm of the Riemann curvature tensor up to a factor.

To see this, let $\Omega_i^j=\frac{\sqrt{-1}}{2\pi} R^j_{i k\overline{\ell}}dz^k\wedge d\overline{z}^\ell$ denote the curvature form, then by Chern-Weil theory the form $Tr(\Omega\wedge\Omega)=\sum_{k,i} \Omega_i^k\wedge\Omega_k^i =\frac{(\sqrt{-1})^2}{4\pi^2}R^k_{ip\overline{q}} R^i_{kr\overline{s}}dz^p\wedge d\overline{z}^q\wedge dz^r\wedge d\overline{z}^s$ represents $c_1^2(M)-2c_2(M)=-2c_2(M)$. You can assume that the metric $\omega$ is the identity at one point so $\omega=\sqrt{-1}\sum_i dz^i\wedge d\overline{z}^i$, and then at that point you compute $$n(n-1)Tr(\Omega\wedge\Omega)\wedge\omega^{n-2}=\sum_{p\neq r}(R^k_{ip\overline{p}} R^i_{kr\overline{r}}-R^k_{ip\overline{r}} R^i_{kr\overline{p}})\omega^n$$ $$=\sum_{p,r}(R^k_{ip\overline{p}} R^i_{kr\overline{r}}-R^k_{ip\overline{r}} R^i_{kr\overline{p}})\omega^n$$ $$=(|\textrm{Ric}|^2-|\textrm{Rm}|^2)\omega^n=- |\textrm{Rm}|^2\omega^n.$$ Integrating this you get what you want, with equality if and only if $\omega$ is flat and so $M$ is finitely covered by a complex torus.

A similar calculation works for any compact Kahler-Einstein manifold, and it can be used to prove the Miyaoka-Yau inequality for manifolds with ample canonical bundle. A reference for a general statement (I think this is not the earliest paper where this result appears) is:

Chen, Bang-yen; Ogiue, Koichi, Some characterizations of complex space forms in terms of Chern classes, Quart. J. Math. Oxford Ser. (2) 26 (1975), no. 104, 459–464.