Negative holomorphic sectional curvature Let X be a complex hermitian manifold with hermitian form $\omega$. How can you prove that if $\omega$ has negative holomorphic sectional curvature, then its scalar curvature is negative, too? 
 A: Here is the answer.
Let $(X,\omega)$ be a Kähler $n$-dimensional manifold. Fix a point $x_0\in X$ an choose local holomorphic coordinates $(z_1,\dots,z_n)$ centered at $x_0$ and such that $(\partial/\partial z_1,\dots,\partial/\partial z_n)$ is unitary at $x_0$. Let
$$
\Theta_{x_0}(T_X,\omega)=\sum_{j,k,l,m=1}^nc_{jklm}\hspace{0.3mm}dz_j\wedge d\bar z_k\otimes\left(\frac\partial{\partial z_l}\right)^*\otimes\frac\partial{\partial z_m}
$$
be the Chern curvature at the point $x_0$. Consider the induced hermitian form on rank one tensors of $T_X\otimes T_X$ given by
$$
\theta_{T_{X,x_o}}(v\otimes w)=\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}v_j\bar v_k w_l\bar w_m,
$$
where
$$
v,w\in T_{X,x_0},\quad v=\sum v_j\hspace{0.3mm}\frac\partial{\partial z_j},\quad w=\sum w_j\hspace{0.3mm}\frac\partial{\partial z_j}.
$$
With this notation, the holomorphic sectional curvature in the direction of $v\in T_{X,x_0}\setminus\{0\}$ is given by 
$$
\frac{1}{||v||_\omega^4}\theta_{T_{X,x_o}}(v\otimes v).
$$
The idea now is to take the average on the $\omega$-unit sphere $S^{2n-1}$ and try to deduce something on the scalar curvature at the point $x_0$ which is given by
$$
s(x_0)=2\sum_{j,k=1}^nc_{jjkk}.
$$ 
So, let's compute the integral
$$
\int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi),
$$
where $d\sigma(\xi)$ is the probability Haar measure on $S^{2n-1}$. It is not hard to see that the integral
$$
\int_{S^{2n-1}}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi)
$$
vanishes unless $j=k$ and $l=m$ or $j=m$ and $k=l$. Thus, we have to compute
$$
\int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi),\quad j,k=1,\dots,n.
$$
It is classically known that
$$
\int_{S^{2n-1}}|\xi_j|^4\hspace{0.3mm}d\sigma(\xi)=\frac 2{n(n+1)},\quad j=1,\dots,n,
$$
and
$$
\int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi)=\frac 1{n(n+1)},\quad 1\le j\ne k\le n.
$$
Then, we get
$$
\begin{aligned}
\int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi) & =\sum_{j,k=1}^nc_{jjkk}\left(\delta_{jk}\frac 2{n(n+1)}+(1-\delta_{jk})\frac 2{n(n+1)}\right) \\
& = \frac 2{n(n+1)}\sum_{j,k=1}^nc_{jjkk}=\frac 1{n(n+1)}s(x_0),
\end{aligned}
$$
where we have used the Kähler identity $c_{jklm}=c_{jmlk}$.
Thus, if $\frac{1}{||v||_\omega^4}\theta_{T_{X,x_o}}(v\otimes v)$ is negative, so is its average and we are done.
