Notation: Let $P_U$ denote the orthogonal projection onto a closed subspace $U$. By $\gamma(T)$ we denote the minimum modulus (lower bound) of a linear operator $T$, i.e., $$ \gamma(T) := \inf\{\|Tx\| : \|x\|=1\}, $$ which is positive iff $T$ has closed range. If $T$ is invertible, then $\gamma(T) = \|T^{-1}\|^{-1}$ (which is the smallest eigenvalue if $T$ is positive definite).
Problem: Let $H$ be a Hilbert space ($H=\mathbb R^n$ is also fine) and let $V$ and $W$ be closed subspaces such that $H = V\dotplus W$, where $\dotplus$ denotes the direct sum. Note that the sum is not necessarily orthogonal. From this alone lots of stuff follows: we have $H = V^\perp\dotplus W^\perp$ and \begin{align} c_0^2 &:= \|P_VP_W\|^2 = \|P_WP_V\|^2 = \|P_V|W\|^2 = \|P_W|V\|^2 = \|P_{W^\perp}|V^\perp\|^2 = \|P_{V^\perp}|W^\perp\|^2\\ &\,\,= 1-\gamma^2(P_{W^\perp}|V) = 1-\gamma^2(P_{V^\perp}|W) = 1-\gamma^2(P_{V}|W^\perp) = 1-\gamma^2(P_{W}|V^\perp)\\ &\,\,<1. \end{align} Now, the operator $Q := P_V+P_W$ is invertible (and thus positive definite). Proof is below. What I would like to have is an estimate for the lower bound $\gamma(Q)$ of $Q$ in terms of $c_0$. I tried a lot but nothing leads to success. I would be happy if anyone could help.
Proof of invertibility of $Q$. Otherwise there would exist a sequence $(x_n)$ with $\|x_n\|=1$ such that $P_Vx_n+P_Wx_n\to 0$. Since $P_V$ and $P_W$ are positive semi-definite, this implies $P_Vx_n\to 0$ and $P_Wx_n\to 0$ and so $(P_V-P_W)x_n\to 0$. But \begin{align} \|P_Vx-P_Wx\|^2 &= \|P_WP_Vx - P_Wx\|^2 + \|P_{W^\perp}P_Vx\|^2\\ &= \|P_WP_{V^\perp}x\|^2 + \|P_{W^\perp}P_Vx\|^2\\ &\ge \gamma^2(P_W|V^\perp)\|P_{V^\perp}x\|^2 + \gamma^2(P_{W^\perp}|V)\|P_Vx\|^2\\ &= (1-c_0^2)\|x\|^2, \end{align} which yields a contradiction.
I've got it! We have \begin{align} (Q-I)^2 &= P_V + P_V(P_W-I) + (P_W-I)P_V + (P_W-I)^2\\ &= P_VP_W + P_WP_V - P_V + P_W - 2P_W + I\\ &= I - (P_V-P_W)^2. \end{align} Since $(P_V-P_W)^2\ge 1-c_0^2$ (see proof above), we get $$ (Q-I)^2\,\le\,1 - (1-c_0^2) = c_0^2. $$ Thus, $-c_0\le Q-I\le c_0$ and therefore $1-c_0\le Q\le 1+c_0$.