I want to prove the following statement (which I heavily believe is true):
Let $g:\mathbb{CP}^n\to \mathbb{R}$ be a non-constant Morse-Bott function and denote by $\text{Icrit}(g)$ the set of isolated critical points of $g$. Then $\#\text{Icrit}(g)\geq n+1$.
So far what I have is the following piece of proof. But first I need to give some technical baggage:
Let $M$ be a closed smooth manifold, define the Poincaré polynomial in $\mathbb{Z}$ to be $$P_t(M)=\sum_i \text{rank}\hspace{1mm} H_i(M;\mathbb{Z})t^i,$$ and, for a given Morse-Bott function $f$ on $M$, let \begin{equation}\label{MBpolynomial} MB_t(f)=\sum_j P_t(M_j)t^{|M_j|}, \end{equation} be the Morse-Bott polynomial. Here the sum runs over the critical submanifolds of $f$ and $|M_j|$ denotes the index of the critical submanifold, that is, the dimension of the negative normal bundle of $M_j$. Then we have the following result often referred to as Morse-Bott inequalities (taken from Liviu Nicolaescu's "An invitation to Morse Theory" first edition, Corollary 2.45)
Let $f:M\rightarrow\mathbb{R}$ be a Morse-Bott function on a compact manifold s.t. the negative normal bundle of every critical submanifold is orientable. Then there exists a polynomial $Q_t$ with non-negative coefficients such that $$MB_t(f)=P_t(M)+(1+t)Q_t.$$
Now, my attempt of proof:
By computing the singular homology of $\mathbb{CP}^n$ we can see that $$P_t(\mathbb{CP}^n)=\sum_{i=0}^nt^{2i}.$$ By dimensional reasons $g$ can only have isolated critical points of index $0,\ldots,2n$; and positive dimensional critical submanifolds of index $0,\ldots,2n-1$, by the nondegeneracy assumption on the critical submanifolds. Denote by $a_0,\ldots,a_{2n}$ the number of isolated critical points of index $0,\ldots,2n$; respectively. Then, by the definition of the Morse-Bott polynomial we have that \begin{equation} \text{MB}_t(g)=\sum_{i=0}^{2n}a_it^i+\sum_jP_t(M_j)t^{|M_j|}, \end{equation} where the $M_j$ denote the (positive dimensional) critical submanifolds and $|M_j|$ their index. On the other hand, by the afore mentioned Theorem \begin{equation} \text{MB}_t(g)=\sum_{i=0}^{n}t^{2i}+(1+t)Q_t, \end{equation} where $Q_t$ is a polynomial with non-negative integer coeficientes. By dimensional reasons we have that $\deg P_t(M_j)t^{|M_j|}\leq 2n$, then the two previous equations imply that $\deg Q_t\leq 2n-1$. Let us say, $$Q_t=\sum_{i=0}^{2n-1}q_it^i.$$ So, we have \begin{equation}\label{MB3} \text{MB}_t(g)=1+q_0+\sum_{i=1}^{n-1}(1+q_{2i}+q_{2i-1})t^{2i}+\sum_{i=0}^{n-1}(q_{2i}+q_{2i+1})t^{2i+1}+(1+q_{2n-1})t^{2n}. \end{equation} Making $t=-1$ in both representations of the Morse-Bott polynomial, we obtain \begin{equation}\label{maket=-1} \sum_{i=0}^na_{2i}=n+1+\sum_{i=0}^{n-1}a_{2i+1}-\sum_jP_{(-1)}(M_j)(-1)^{|M_j|}. \end{equation} Then we can see that, if the following holds, then my statement is proven: \begin{equation} 2\sum_{i=0}^{n-1}a_{2i+1}-\sum_jP_{(-1)}(M_j)(-1)^{|M_j|}\geq 0. \end{equation}
But this last claim is what I'm unable to prove. I would appreciate any help I can get.