Let $\mathfrak{g}$ be a semi-simple finite dimensional Lie algebra.
Denote by $L(\lambda)$ an irreducible finite-dimensional $\mathfrak{g}$-module of highest weight $\lambda$. (I.e. $\lambda$ is integral dominant weight)
Suppose that the multiplicity of $L(\nu)$ in the tensor product $L(\lambda)\otimes L(\mu)$ is greater than $0$. 


Does this imply that either $\lambda \geq \nu$ or $\lambda \leq \nu$?
Where the comparison is the standard partial ordering on the set of weights:
$\lambda \geq \nu$ iff their difference $\lambda-\nu$ is equal to the sum of positive roots with nonnegative coefficients.



Does the nonvanishing of a Littlewood-Richardson coefficient implies comparability of highest weights?
What you want is not true. For example, take $\mathfrak{sl}\_3$ and $\lambda = \mu = \omega\_1$, the first dominant weight. Then the multiplicity of $\nu = \omega\_2$ is $1$. But 
$$
\nu - \lambda = \omega\_2 - \omega\_1 = \frac13 \alpha\_2 - \frac13 \alpha\_1,
$$
so it is neither negative, nor positive.


On the other hand, it is known that $\nu$ is in the convex hull of the weights $\lambda + w\mu$, where $w$ runs over the Weyl group. On the other hand $\mu - w\mu \ge 0$ with respect to your ordering. It follows that 
$$
\nu \le \lambda + \mu.
$$
Analogously one can check that $\nu \ge \lambda + w\_0\mu$, where $w\_0$ is the longest element in the Weyl group.


