I am not sure about the history of this result. You can find in Thierry Aubin's book "Some nonlinear problems in Riemannian geometry", Theorem 4.20. He attributes it to his own 1978 paper.

The proof is really simple, though. If you write $\Delta f=g^{i\overline{j}}f_{i\overline{j}}$ for the complex Laplacian, and assume that $\Delta f=-\lambda f$ with $\lambda$ the first eigenvalue of $\Delta$ (which is half of the first eigenvalue of the real Laplacian), then commuting covariant derivatives you can easily see that $$\int_M (\Delta f)^2=\int_M g^{i\overline{j}}g^{p\overline{q}}f_{i\overline{j}} f_{\overline{q}p}=-\int_M g^{i\overline{j}}g^{p\overline{q}}f_{i} f_{\overline{q}p\overline{j}}=-\int_M g^{i\overline{j}}g^{p\overline{q}}f_{i} f_{\overline{q}\overline{j}p} +\int_M R^{i\overline{j}} f_i f_{\overline{j}}$$ $$=\int_M |f_{ij}|^2+\int_M R^{i\overline{j}} f_i f_{\overline{j}}.$$ If you assume that $\mathrm{Ric}\geq k>0$, then $R_{i\overline{j}}\geq kg_{i\overline{j}}$ and so $$\lambda^2\int_M f^2 =\int_M(\Delta f)^2\geq k\int_M |\partial f|^2=k\lambda\int_M f^2,$$ and so $\lambda\geq k$.

I am not sure about the history of this result. You can find in Thierry Aubin's book "Some nonlinear problems in Riemannian geometry", Theorem 4.20. He attributes it to his own 1978 paper.

The proof is really simple, though. If you write $\Delta f=g^{i\overline{j}}f_{i\overline{j}}$ for the complex Laplacian, and assume that $\Delta f=-\lambda f$ with $\lambda$ the first eigenvalue of $\Delta$ (which is half of the first eigenvalue of the real Laplacian), then commuting covariant derivatives you can easily see that $$\int_M (\Delta f)^2=\int_M g^{i\overline{j}}g^{p\overline{q}}f_{i\overline{j}} f_{\overline{q}p}=-\int_M g^{i\overline{j}}g^{p\overline{q}}f_{i} f_{\overline{q}p\overline{j}}=-\int_M g^{i\overline{j}}g^{p\overline{q}}f_{i} f_{\overline{q}\overline{j}p} +\int_M R^{i\overline{j}} f_i f_{\overline{j}}$$ $$=\int_M |f_{ij}|^2+\int_M R^{i\overline{j}} f_i f_{\overline{j}}.$$ If you assume that $\mathrm{Ric}\geq k>0$, then $R_{i\overline{j}}\geq kg_{i\overline{j}}$ and so $$\lambda^2\int_M f^2 =\int_M(\Delta f)^2\geq k\int_M |\partial f|^2=k\lambda\int_M f^2,$$ and so $\lambda\geq k$.

I am not sure about the history of this result. You can find in Thierry Aubin's book "Some nonlinear problems in Riemannian geometry", Theorem 4.20. He attributes it to his own 1978 paper.

The proof is really simple, though. If you write $\Delta f=g^{i\overline{j}}f_{i\overline{j}}$ for the complex Laplacian, and assume that $\Delta f=-\lambda f$ with $\lambda$ the first eigenvalue of $\Delta$ (which is half of the first eigenvalue of the real Laplacian), then commuting covariant derivatives you can easily see that $$\int_M (\Delta f)^2 -\int_M |f_{ij}|^2=\int_M R^{i\overline{j}} f_i f_{\overline{j}}.$$ If you assume that $\mathrm{Ric}\geq k>0$, then $R_{i\overline{j}}\geq kg_{i\overline{j}}$ and so $$\lambda^2\int_M f^2 =\int_M(\Delta f)^2\geq k\int_M |\partial f|^2=k\lambda\int_M f^2,$$ and so $\lambda\geq k$.

I am not sure about the history of this result. You can find in Thierry Aubin's book "Some nonlinear problems in Riemannian geometry", Theorem 4.20. He attributes it to his own 1978 paper.

The proof is really simple, though. If you write $\Delta f=g^{i\overline{j}}f_{i\overline{j}}$ for the complex Laplacian, and assume that $\Delta f=-\lambda f$ with $\lambda$ the first eigenvalue of $\Delta$ (which is half of the first eigenvalue of the real Laplacian), then commuting covariant derivatives you can easily see that $$\int_M (\Delta f)^2=\int_M g^{i\overline{j}}g^{p\overline{q}}f_{i\overline{j}} f_{\overline{q}p}=-\int_M g^{i\overline{j}}g^{p\overline{q}}f_{i} f_{\overline{q}p\overline{j}}=-\int_M g^{i\overline{j}}g^{p\overline{q}}f_{i} f_{\overline{q}\overline{j}p} +\int_M R^{i\overline{j}} f_i f_{\overline{j}}$$ $$=\int_M |f_{ij}|^2+\int_M R^{i\overline{j}} f_i f_{\overline{j}}.$$ If you assume that $\mathrm{Ric}\geq k>0$, then $R_{i\overline{j}}\geq kg_{i\overline{j}}$ and so $$\lambda^2\int_M f^2 =\int_M(\Delta f)^2\geq k\int_M |\partial f|^2=k\lambda\int_M f^2,$$ and so $\lambda\geq k$.