Let $X$ be a smooth Calabi-Yau 3-fold with Kahler form $w$,
It is true that $\int c_2(TX) \wedge w \geq 0$ (for any Kahler form $w$ on $X$).
Proof via algebraic geometry is rather difficult. Some where It was saying that for such $X$
$$\int c_2(TX) \wedge w = \int \left\| R \right\|^2 dvol $$
where $R$ is the curvature tensor; so it is non negative and zero only if $X$ is abelian variety .
Can any body write down in local coordinates or ...(or refer to somewhere) why the identity above is true.
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.
The reason is that, for a Calabi-Yau $3$-fold, the curvature tensor $R$ is an irreducible representation of $SU(3)$. Since the form representing the Chern class $c_2(TX)$, when computed with respect to the Calabi-Yau metric, is clearly quadratic in the curvature tensor and since $c_2(TX)\wedge\omega$ is a $(3,3)$-form that is a quadratic form in the curvature times the metric volume form, the irreducibility of the curvature implies that there is a universal constant $\lambda$ such that $c_2(TX)\wedge\omega = \lambda\ \|R\|^2\ dvol$ for all Calabi-Yau $3$-folds. Checking on a single example (say, a quintic hypersurface in $\mathbb{CP}^4$) shows that $\lambda >0$.
The irreducibility comes from the fact (due, I believe, to Bochner) that, for a general Kähler metric on an $n$-fold, the curvature tensor breaks up into $3$ irreducible parts under $U(n)$: the scalar curvature, the traceless Ricci curvature, and the Bochner curvature. For a Calabi-Yau metric, the first two vanish. Thus, the above formula actually generalizes to $c_2(TX)\wedge\omega^{n-2} = \lambda_n\ \|R\|^2\ dvol$ for all Calabi-Yau $n$-folds, where $\lambda_n>0$ for all $n\ge 2$.
adding my comment as an answer.
We have the following Kobayashi identity $$(∣∣\frac{i}{2\pi}Θ_ω∣∣^2−|Ric \omega|^2)\frac{ω^n}{n!}=(2c_2−c_1^2)∧\frac{ω^{n−2}}{(n−2)!}$$ where $\frac{i}{2π}Θ_ω$ is the curvature tensor of $ω$ and $Ric(ω)$ is the Ricci-form of $ω$ so since $X$ is Calabi-Yau variety , hence we have Yau's theorem (Ricci flat)and $c_1=0$ by integrating in both parts and take $n=3$ , we obtain your identity, see the book of complex vector bundle of Kobayashi
