Let $\rho$ and $\sigma$ be two irreducible representations of the same simple Lie algebra $\mathfrak{g}$. Under which conditions does the decomposition of the tensor product $\rho \otimes \sigma$ into irreducible representations contain the adjoint representation of $\mathfrak{g}$?
My guess is that $\rho$ and $\sigma$ must be dual. But is this true? And if yes, how could one proof it?
As the comments suggest, there don't seem to be any easily stated necessary and sufficient conditions for the adjoint representation to occur as a summand of the tensor product of two irreducibles (say with highest weights $\lambda$ and $\mu$). Of course, one has to fix a simple system of roots to speak of "highest weight".
However, there is a simple necessary condition. Start with the classical fact that 0 is a weight of an irreducible representation precisely when the highest weight lies in the root lattice. In particular, this applies to the adjoint representation, whose highest weight is the highest root; here the 0 weight space corresponds to a Cartan subalgebra. In turn, classical results show that the highest weights $\nu$ of all irreducible summands lie below the sum of the two given highest weights in the usual partial ordering of weights. It follows immediately that the adjoint representation occurs as a summand of the tensor product only if $\lambda + \mu$ lies in the root lattice.
In some Lie types such as $G_2$, the root lattice equals the weight lattice. But here one sees readily that not all tensor products of two irreducibles have the 14-dimensional adjoint representation as a summand. I don't know of an easily stated sufficient condition in general.
