= Solution
The <Trace form of a Lie algebra representation> is invariant because
$$
\begin{aligned}
H([X,Y],Z)
&=\operatorname{Tr}([d(X),d(Y)]d(Z))\\
&=\operatorname{Tr}(d(X)[d(Y),d(Z)])
=H(X,[Y,Z]).
\end{aligned}
$$
Choose a basis orthonormal for the positive-definite form $-\kappa$ of the compact simple algebra. Any invariant bilinear form determines an endomorphism commuting with the irreducible adjoint action; <Schur lemma> makes it scalar. Thus $H(T_a,T_b)=c\delta_{ab}$. Since $d(T_a)$ is anti-Hermitian,
$$
H(T_a,T_a)=-\operatorname{Tr}(d(T_a)^\dagger d(T_a))<0.
$$
The inequality is strict because the kernel of the nontrivial irreducible representation is an ideal and hence zero. Therefore $c=-\mu$ with $\mu>0$.
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