[proofplan]
Write the positive root $\gamma$ in the simple-root basis and use its non-simplicity to see that its height is at least $2$. If every simple root occurring in this expansion had non-positive inner product with $\gamma$, then positive definiteness would force $(\gamma,\gamma) \leq 0$, impossible for a non-zero root. Hence some simple root $\alpha$ occurring in $\gamma$ satisfies $(\gamma,\alpha)>0$; the root-string property then implies that $\gamma-\alpha$ is a root, and its simple-root expansion has non-negative coefficients, so it is positive.
[/proofplan]
[step:Expand $\gamma$ in the simple-root basis and isolate the roots that occur]
Since $\gamma \in \Phi^+$, there are uniquely determined integers $n_\beta \in \mathbb{Z}_{\geq 0}$, indexed by $\beta \in \Delta$, such that
\begin{align*}
\gamma = \sum_{\beta \in \Delta} n_\beta \beta.
\end{align*}
Define the support of this expansion by
\begin{align*}
S := \{\beta \in \Delta : n_\beta > 0\}.
\end{align*}
Because $\gamma$ is a root, $\gamma \neq 0$, so $S \neq \varnothing$. Since $\gamma \notin \Delta$, the expansion is not the expansion of a single simple root with coefficient $1$; equivalently,
\begin{align*}
\operatorname{ht}(\gamma) := \sum_{\beta \in \Delta} n_\beta \geq 2.
\end{align*}
[/step]
[step:Find a simple root in the support with positive inner product against $\gamma$]
Assume, toward a contradiction, that
\begin{align*}
(\gamma,\beta) \leq 0
\end{align*}
for every $\beta \in S$. Using the expansion of $\gamma$ and bilinearity of the inner product, we obtain
\begin{align*}
(\gamma,\gamma)
&= \left(\gamma,\sum_{\beta \in \Delta} n_\beta \beta\right) \\
&= \sum_{\beta \in \Delta} n_\beta(\gamma,\beta) \\
&= \sum_{\beta \in S} n_\beta(\gamma,\beta) \leq 0.
\end{align*}
This contradicts positive definiteness of the Euclidean inner product, since $\gamma \neq 0$ implies $(\gamma,\gamma)>0$. Therefore there exists $\alpha \in S$ such that
\begin{align*}
(\gamma,\alpha) > 0.
\end{align*}
[guided]
We need to find a simple root that can be subtracted from $\gamma$. The only simple roots that can be subtracted without immediately producing a negative coefficient are the simple roots in the support
\begin{align*}
S := \{\beta \in \Delta : n_\beta > 0\}.
\end{align*}
So we prove that at least one of these roots has positive inner product with $\gamma$.
Suppose the opposite: for every $\beta \in S$, we have $(\gamma,\beta) \leq 0$. Since $\gamma$ has the simple-root expansion
\begin{align*}
\gamma = \sum_{\beta \in \Delta} n_\beta \beta,
\end{align*}
bilinearity of the inner product gives
\begin{align*}
(\gamma,\gamma)
&= \left(\gamma,\sum_{\beta \in \Delta} n_\beta \beta\right) \\
&= \sum_{\beta \in \Delta} n_\beta(\gamma,\beta).
\end{align*}
Terms with $\beta \notin S$ vanish because $n_\beta = 0$, so
\begin{align*}
(\gamma,\gamma)
= \sum_{\beta \in S} n_\beta(\gamma,\beta).
\end{align*}
For each $\beta \in S$, the coefficient $n_\beta$ is positive and the factor $(\gamma,\beta)$ is non-positive by assumption. Therefore every summand is non-positive, and hence
\begin{align*}
(\gamma,\gamma) \leq 0.
\end{align*}
But $\gamma$ is a non-zero vector in the Euclidean space $E$, so positive definiteness gives $(\gamma,\gamma)>0$. This contradiction proves that there is some $\alpha \in S$ satisfying
\begin{align*}
(\gamma,\alpha)>0.
\end{align*}
[/guided]
[/step]
[step:Use the root-string property to subtract this simple root]
Apply the root-string property to the roots $\gamma,\alpha \in \Phi$. It says that there exist integers $p,q \in \mathbb{Z}_{\geq 0}$ such that
\begin{align*}
\{\gamma + k\alpha : k \in \mathbb{Z}\} \cap \Phi
=
\{\gamma - p\alpha,\gamma-(p-1)\alpha,\dots,\gamma+q\alpha\}
\end{align*}
and
\begin{align*}
p-q = \frac{2(\gamma,\alpha)}{(\alpha,\alpha)}.
\end{align*}
Here $(\alpha,\alpha)>0$ because $\alpha \neq 0$, and $(\gamma,\alpha)>0$ by the previous step. Hence
\begin{align*}
p-q > 0.
\end{align*}
Since $q \geq 0$, this implies $p \geq 1$. Therefore the first negative-direction element of the string exists, so
\begin{align*}
\gamma-\alpha \in \Phi.
\end{align*}
(citing a result not yet in the wiki: Root String Property)
[guided]
Now we use the standard root-string mechanism. For two roots $\gamma,\alpha \in \Phi$, the root-string property says that the roots on the affine line $\gamma+\mathbb{Z}\alpha$ form one consecutive string:
\begin{align*}
\{\gamma + k\alpha : k \in \mathbb{Z}\} \cap \Phi
=
\{\gamma - p\alpha,\gamma-(p-1)\alpha,\dots,\gamma+q\alpha\},
\end{align*}
where $p,q \in \mathbb{Z}_{\geq 0}$, and the string lengths satisfy
\begin{align*}
p-q = \frac{2(\gamma,\alpha)}{(\alpha,\alpha)}.
\end{align*}
The denominator is positive because $\alpha$ is a non-zero vector in the Euclidean space $E$, so $(\alpha,\alpha)>0$. From the previous step, $(\gamma,\alpha)>0$. Therefore
\begin{align*}
\frac{2(\gamma,\alpha)}{(\alpha,\alpha)} > 0,
\end{align*}
and hence
\begin{align*}
p-q > 0.
\end{align*}
Because $q$ is a non-negative integer, $p-q>0$ forces $p \geq 1$. The inequality $p \geq 1$ means precisely that the string contains the root one step in the negative $\alpha$-direction from $\gamma$. Thus
\begin{align*}
\gamma-\alpha \in \Phi.
\end{align*}
(citing a result not yet in the wiki: Root String Property)
[/guided]
[/step]
[step:Check that the resulting root is positive]
Because $\alpha \in S$, we have $n_\alpha \geq 1$. Subtracting $\alpha$ from the simple-root expansion of $\gamma$ gives
\begin{align*}
\gamma-\alpha
=
(n_\alpha-1)\alpha + \sum_{\substack{\beta \in \Delta \\ \beta \neq \alpha}} n_\beta \beta.
\end{align*}
All coefficients in this expansion are non-negative integers. Since the previous step proves $\gamma-\alpha \in \Phi$, the defining decomposition $\Phi = \Phi^+ \sqcup (-\Phi^+)$ determined by $\Delta$ implies
\begin{align*}
\gamma-\alpha \in \Phi^+.
\end{align*}
Thus there exists $\alpha \in \Delta$ such that $\gamma-\alpha$ is a positive root, as required.
[/step]
