Lie algebras and roots systems Let $\Phi$ an irreducible system of roots, $\Phi^{+} \subset \Phi$ a choose of positive roots. I have to prove that if $(\alpha, \beta) \ge 0$ for al $\beta \in \Phi^{+}$ then $\alpha$ is the highest among roots of the same lenght. I have a long way... is there a way that uses only some theorem?
My way: Let be $\alpha '\neq \alpha$ the root of maximal lenght $||\alpha||$. Obviusly $\alpha '\in \Phi_+$, so $(\alpha,\alpha')\ge 0$. \
I want to prove that $(\alpha,\alpha')>0$. \
I show now that
$$\alpha'=\sum_{\alpha_i\in\Delta}m_i\alpha_i\,\,\,\,\,\,\,\,(m_j>0)$$
We can suppose that $\alpha'$ is short.
Now we suppose that $\alpha'$ is short. $\Phi$ is irreducible than it exists a simple root $\beta_i$ such that
$$(\alpha',\beta_i)<0$$
so $\alpha'+\beta_i$ is a positive root longer than $\alpha'$.\ Morever $||\alpha'+\beta_i||=||\alpha'||$, because
$$(\alpha'+\beta_i,\alpha'+\beta_i)=(\alpha',\alpha')+(\beta_i,\beta_i)+2(\alpha'+\beta_i)=$$
$$ =(\alpha',\alpha')+(\beta_i,\beta_i)\left[1+\frac{2(\alpha',\beta_i)}{(\beta_i,\beta_i)} \right] $$
$$\leq (\alpha',\alpha')$$
Because we have $<\alpha',\beta_i>\leq -1$, for the irreducibility of $\Phi$ we have finish.\
Now if $(\alpha,\alpha')=0$, we have
$$\sum m_j(\alpha,\alpha_j)=0 \,\,\,\,\,\,\,\,\,\,\, (\alpha_j\in\Delta)$$
and so
$$(\alpha,\alpha_j)=0 \,\,\,\,\,\,\,\,\,\,\,\,\, (\forall\,\alpha_j\in\Delta)$$
and it is no possible.\
We have finally that $(\alpha',\alpha)>0$, so $\alpha'-\alpha \in \Phi_+$,because $\alpha maximality$.
However we had seen $(\alpha',\alpha)>0$ and $(\alpha',\alpha')/(\alpha,\alpha)=1$ we must have $<\alpha',\alpha>=1$. So we can say that
$$(\alpha',\alpha)=\frac{1}{2}(\alpha,\alpha) $$
but for hipotesys we have to have
$$(\alpha,\alpha'-\alpha)\ge 0 $$
and so
$$(\alpha,\alpha')-(\alpha,\alpha)=-\frac{1}{2}(\alpha,\alpha)\ge 0 $$
and it si no possible.
Could you give me a more easy and fast method... maybe using theorems? Is there a method that involves Weyl group?
 A: Let $\alpha_0$ be the highest root of the same length as $\alpha$. Then $\alpha$ and $\alpha_0$ are in the same orbit of the Weyl group. Therefore (are you familiar with the length function on the Weyl group?) there is a sequence of simple reflections $s_{i_1}$,
$s_{i_2}$, $\ldots$,$s_{i_\ell}$ such that
$$
\alpha=s_{i_\ell}s_{i_{\ell-1}}\cdots s_{i_1}(\alpha_0).
$$
Furthermore, by selecting the minimal number (= minimal $\ell$) of such simple reflections we have the result that the recursively defined sequence of roots $\alpha_1=s_{i_1}(\alpha_0)$,
$\alpha_2=s_{i_2}(\alpha_1)$, $\ldots$, $\alpha=\alpha_\ell=s_{i_\ell}(\alpha_{\ell-1})$ is linearly ordered:
$$
\alpha<\alpha_{\ell-1}<\alpha_{\ell-2}<\cdots <\alpha_0.
$$ 
So if here $\alpha\neq\alpha_0$, or equivalently $\ell>0$, then the simple root $\beta_{i_\ell}$ corresponding to the simple reflection $s_{i_\ell}$ satisfies $(\alpha,\beta_{i_\ell})<0$. This contradicts your hypothesis.
This type of arguments are common, when describing the theory leading to the length of elements of the Weyl group. If you haven't covered that yet, then it gets messier.
