Proof that higher genus surface admits a metric of negative Ricci scalar everywhere In the Green, Schwarz and Witten Superstring Theory textbook, the paragraph below equation 3.3.15 says, 

For genus greater than one, it can be shown that the surface admits a metric of everywhere negative scalar curvature.

This statement is essential in the proof that $C_g$ for $g>1$ is always zero, but I cannot prove that. Could you point out how to prove this?
 A: This result follows from i) Uniformization theorem and ii) Gauss-Bonnet theorem in 2d. 
According to the statement of uniformization theorem from this wiki page : 

every connected Riemann surface X admits a unique complete 2-dimensional real Riemann metric with constant curvature −1, 0 or 1 inducing the same conformal structure

On the other hand, according to Gauss-Bonnet theorem, integral of the scalar curvature on a 2d surface is a positive multiple of the Euler characteristic ($\chi=2-2g$). Since the Euler characteristic is negative for $g>1$ so from the uniformization theorem it follows that -

any Riemann surface with genus $g>1$ admits a unique complete 2-dimensional real Riemann metric with constant curvature −1 inducing the same conformal structure


Note : I am not aware of any good reference where uniformization thereom is proved in the form stated above. However, I hope you can find a proof in some of the references mentioned in the corresponding wiki article. 
A: Why not use explicit construction for such a surface?
From The Manifold Atlas:

Any hyperbolic metric on a closed, orientable surface $S_g$ of genus $g\ge 2$ is obtained by the following 
  construction: choose a geodesic $4g$-gon in the hyperbolic plane ${\Bbb H}^2$ with area $4(g-1)\pi$. (This implies that the sum of interior angles is $2\pi$.) Then choose orientation-preserving isometries $I_1,J_1,\ldots,I_g,J_g$ which realise the gluing pattern of $S_g$: for $j=1,\ldots,g$ we require that $I_j$ maps $a_j$ to $\overline{a}_j$, $J_j$ maps $b_j$ to $\overline{b}_j$. Let $\Gamma\subset Isom^+\left({\Bbb H}^2\right)$ be the subgroup generated by $I_1,J_1,\ldots,I_g,J_g$. Then $\Gamma$ is a discrete subgroup of $Isom^+\left({\Bbb H}^2\right)$ and $\Gamma\backslash{\Bbb H}^2$ is a hyperbolic surface diffeomorphic to ${\Bbb H}^2$.

Here is an illustration for genus 2 surface glued from an octagon in hyperbolic plane
(image taken from here):

