Cfrgtkky

Let $Phi$ a root system and $W$ the Weyl group relative to basis $Delta$. Let $mu in overline{C(Delta)}$, where $C(Delta)$ is the Weyl chamber fundamental, $w in W$ and $alpha in Delta$ such that $walpha prec 0$. Show that $(mu, walpha) leq 0$.

Comments:
I need this to understand a demonstration of the Lemma 10.3B of book J.E. Humphreys - Introduction to Lie algebras and representation theory. He states that this fact occurs because $mu in overline{C(Delta)}$. But I can not see why.

I'm assuming that $(mu, walpha) > 0$. If $mu in C(Delta)$ then $walpha in Delta Rightarrow w alpha succ 0$ a contradiction.
I can not solve the case where $mu in overline{C(Delta)} setminus C(Delta)$.

Another way is as follows: $(mu, w alpha) leq 0 Leftrightarrow -(mu, w alpha) geq 0 Leftrightarrow (mu, w(- alpha)) geq 0 Leftrightarrow w(- alpha) in Delta$. Knowing that $alpha in Delta$ and $walpha prec 0$
I can conclude that $w(-alpha) in Delta$?

Write $alpha_1,dots,alpha_n$ for the given choice of simple roots. Suppose $beta$ is a root with $beta prec 0$, that is
$$beta=sum_{i=1}^n k_i alpha_i,$$ with $k_i leq 0$. We must show that $(beta,mu)$ is non-positive for all $mu$ in the closure of the fundamental chamber. But this is an immediate consequence of the definition, which we now recall.

The fundamental chamber $C(Delta)$ is defined by
$$C(Delta)={v in mathfrak{h} | (alpha,v) > 0 quad hbox{for all positive roots $alpha$} }$$ and its closure is
$$overline{C(Delta)}={v in mathfrak{h} | (alpha,v) geq 0 quad hbox{for all positive roots $alpha$} }.$$ Therefore by definition, for any $mu in overline{C(Delta)}$,
$$(beta,mu)=sum_{i=1}^n k_i (alpha_i,mu) leq 0.$$