What does the Kähler cone of the one-point blow-up of $\mathbb{C}P^n$ look like? I found that related to the Kähler cone there are many discussions on MathOverflow. 
Recently I am interested in the very special manifold of the one-point blow up of $\mathbb{C}P^n$ and just want to see what the general results on Fano Kähler manifolds look like when it comes to this special manifold.
My question is very concrete. Let $x$ and $y$ be the two generators of $H^2(\mathbb{C}P^n\sharp\bar{\mathbb{C}P^n};\mathbb{Z})$ corresponding to the two components respectively. Moreover we assume $\int x^n=-\int y^n=1$. $H^2(\mathbb{C}P^n\sharp\bar{\mathbb{C}P^n};\mathbb{R})=H^{1,1}(\mathbb{C}P^n\sharp\bar{\mathbb{C}P^n};\mathbb{R})$ as $b_2(\mathbb{C}P^n\sharp\bar{\mathbb{C}P^n};\mathbb{R})=2$. Thus every element in $H^{1,1}(\mathbb{C}P^n\sharp\bar{\mathbb{C}P^n};\mathbb{R})$ can be written in the form $ax+by$, $a,b\in\mathbb{R}$. So my question is $$
\{(a,b)\in\mathbb{R}^2~|~ax+by>0\}=?$$ Of course
this set is contained in $\{(a,b)~|~a^n-b^n>0\}$. 
 A: The Kahler cone of any compact manifold is described by a theorem of Demailly and Paun. If $X$ is a compact Kahler manifold, then its Kahler cone is one of the connected components of the set
$$
\mathcal P = 
\lbrace \alpha \in H^{1,1}(X,\mathbb R) \mid
\int_Z \alpha^p > 0 \rbrace
$$
where $Z$ runs through all the $p$-dimensional closed complex subspaces of $X$. If $X$ is projective, then the Kahler cone is actually this set.
Since the blowup of a projective variety in a point is projective, and the cohomology ring of $\mathbb P^n$ blown up in a point is pretty explicit, this lets us calculate the Kahler cone. Indeed, by some fun manipulations one gets
$$
\mathcal P \simeq \lbrace 
(a,b) \in \mathbb R^2 \mid a > 0, \quad b > 0, \quad a > b
\rbrace
$$
where $(a,b) \mapsto aH - bE$. Here $H$ is the divisor of a general hyperplane in $\mathbb P^n$, pulled back to the blowup, and $E$ is the exceptional divisor of the blowup. You seem to be missing the $a, b > 0$ conditions, since $aH - bE > 0$ is equivalent to $a^n - b^n = (a-b) \cdot (a^{n-1} + \ldots + b^{n-1}) > 0$, and one can fulfill this condition with zero or negative $a$ or $b$, which would place us outside of the nef, or even pseudoeffective, cones.
