Ricci curvature of sum of metrics Is there an estimate for Ric$(g+h)$ in terms of Ric$(g)$ and Ric$(h)$, where $g,h$ are smooth Riemannian metrics?
More specifically can one say that the eigenvalues will decrease (resp. increase) if we take h to be something of constant negative (resp. positive) Ricci?
In the examples I have computed this seems to be the case but I am at a loss of how to prove something in general. If it is helpful I can take the metrics to be K\"ahler?
 A: Since you provided me with the escape hatch of only focusing on Kahler metrics, I'll immediately jump out that door without even checking if I have a parachute.
There's a little trick to dealing with curvature tensors of sums of metrics that I'm not sure is very well known. I saw it stated as an exercise in Zheng's Complex differential geometry. We start by considering a short exact sequence of holomorphic vector bundles
$$
0 \to S \to E \to Q \to 0
$$
over a complex manifold of dimension $n$
and recalling the Codazzi-Griffiths equations for the curvatures of the metrics on $S$ and $Q$ induced by a metric on $E$. Then we notice that given a complex manifold $X$ with metrics $h_1$ and $h_2$, there is a short exact sequence
$$
0 \to T_X \to T_X \oplus T_X \to T_X \to 0,
$$
where the first nonzero arrow is $v \mapsto (v,v)$, the second one is $(v,w) \mapsto v-w$, the metric on the direct sum is $h_1 \oplus h_2$, the one on the subbundle is $h_1 + h_2$, and the "quotient" bundle gets the induced metric. Now we just apply Codazzi-Griffiths. (This can be done for any holomorphic vector bundle. Also works in the Riemannian case, actually.)
This yields that the curvature tensor of $h_1+h_2$ is
$$
R(x,\bar y, z, \bar w)
= R_1(x, \bar y, z,\bar w)
+ R_2(x, \bar y, z,\bar w)
- \langle b(x)(z), \overline{b(y)(w)}\rangle_Q,
$$
where $b(x)(z) := (D_{h_1,x} - D_{h_2,x})z$ is the second fundamental form and $h_Q$ is the metric on the "quotient".
To get from this to the Ricci tensor, we're going to integrate two of the tensor elements over the unit sphere with respect to $h_1 + h_2$ (which is the same as taking the trace with respect to them up to a constant factor). The calculations are routine but a bit long, so I'll just say that if we set $m := \min |v|_{h_1}$ and $M := \max |v|_{h_1}$, where $v$ ranges over the unit sphere according to $h_1 + h_2$, then the Ricci tensor of $h_1 + h_2$ is
$$
r(x, \bar y)
= \frac{{M^2}^{2n} - {m^2}^{2n}}{2(n+1)} r_1(x, \bar y)
+ \frac{{(1-m^2)}^{2n} - {(1-M^2)}^{2n}}{2(n+1)} r_2(x, \bar y)
- h_{\operatorname{End}(T_X)}(b(x), \overline{b(y)}),
$$
where $h_{\operatorname{End}(T_X)}$ is the metric we get on $\operatorname{End}(T_X) = \operatorname{Hom}(T_X, T_X)$ by equipping the first factor with $h_1+h_2$ and the second factor with $h_Q$ and calculating the usual Hilbert-Schmidt metric.
You can do essentially the same thing in the Riemannian setting and get something involving more terms with the second fundamental form, see for example Cavenaghi and Sperança.
Now, is this helpful? That, uhh, remains to be seen. To be able to answer your question on the eigenvalues you have to be able to estimate the contribution of the metric on the endomorphism bundle. That involves the metric on the "quotient", which can be explicitly calculated (a fun exercise in linear algebra), but honestly I've never been happier after looking at it than I was before.
