This is false. There exists a smooth projective surface $X$ with a strictly nef divisor $D$ (so $D\cdot C>0$ for all curves $C\subset X$) and yet $(D^2)=0$, so in particular $D$ is not ample, see e.g. here.

Letting $\alpha=c_1(D)$ we have $\int_C \alpha>0$ for all curves $C$. Since $\alpha$ is nef, for any Kähler form $\omega$ we have $\int_X \alpha\wedge\omega\geq 0.$ And if this integral was equal to zero, this would force $\alpha$ to be the zero class, which is not the case.

This is false. There exists a smooth projective surface $X$ with a strictly nef divisor $D$ (so $D\cdot C>0$ for all curves $C\subset X$) and yet $(D^2)=0$, so in particular $D$ is not ample, see e.g. here.

Letting $\alpha=c_1(D)$ we have $\int_C \alpha>0$ for all curves $C$. Since $\alpha$ is nef, for any Kähler form $\omega$ we have $\int_X \alpha\wedge\omega\geq 0.$ And if this integral was equal to zero, this would force $\alpha$ to be the zero class (by the Hodge index theorem), which is not the case.

This is false. There exists a smooth projective surface $X$ with a strictly nef divisor $D$ (so $D\cdot C>0$ for all curves $C\subset X$) and yet $(C^2)=0$, so in particular $D$ is not ample, see e.g. here.

Letting $\alpha=c_1(D)$ we have $\int_C \alpha>0$ for all curves $C$. Since $\alpha$ is nef, for any Kähler form $\omega$ we have $\int_X \alpha\wedge\omega\geq 0.$ And if this integral was equal to zero, this would force $\alpha$ to be the zero class, which is not the case.

This is false. There exists a smooth projective surface $X$ with a strictly nef divisor $D$ (so $D\cdot C>0$ for all curves $C\subset X$) and yet $(D^2)=0$, so in particular $D$ is not ample, see e.g. here.

Letting $\alpha=c_1(D)$ we have $\int_C \alpha>0$ for all curves $C$. Since $\alpha$ is nef, for any Kähler form $\omega$ we have $\int_X \alpha\wedge\omega\geq 0.$ And if this integral was equal to zero, this would force $\alpha$ to be the zero class, which is not the case.