Is it possible for a Lagrangian to give equations of motion which are not consistent? Consider a Lagrangian $\mathcal{L}$ which is function of, for example, some vector fields $A^\mu$ and tensor fields $B^{\mu\nu}$. That is,
\begin{align}
\mathcal{L}=\mathcal{L}(A^\mu, B^{\mu\nu})
\end{align}
Then I would like to ask that is it possible to derive from such $\mathcal{L}$ different equations of motion which are not consistent by using Euler-Lagrange equation but with regard to different fields?
By "consistent" I mean for example one equation of motion gives
\begin{align}
A^{\mu}A^{\nu}+B^{\mu\nu}=0
\end{align}
while the other equation of motion writes
\begin{align}
A^{\mu}A^{\nu}+2B^{\mu\nu}=0
\end{align}
If it is possible, then what does it mean? Does it mean that such Lagrangian is a taboo in constructing?
 A: Classical mechanics answer
A simple example of this would be to impose constraints that can't be solved simultaneously, using Lagrange multipliers. For example, let's take a particle in 2 spatial dimensions and require that it is simultaneously on a circle of radius $R$ and a circle of radius $2R$
\begin{equation}
L = \frac{m}{2} \dot{\vec r}^2 + \lambda_1(|\vec{r}|-R) + \lambda_2(|\vec{r}|-2R)
\end{equation}
where $\lambda_1,\lambda_2$ are Lagrange multipliers.
I think the interpretation is that your most fruitful course of action is to give up on this Lagrangian and try again.
I think this answers the question that was asked, but I have added a new section below to clarify what happens quantum mechanically based on some interesting discussion in the comments.
What happens quantum mechanically?
We can construct a path integral from this lagrangian as
\begin{equation}
Z[J] = \int \mathcal{D} \vec{r} \mathcal{D} \lambda_1 \mathcal{D} \lambda_2 e^{i \left(\frac{m}{2} \dot{\vec r}^2 + \lambda_1(|\vec{r}|-R) + \lambda_2(|\vec{r}|-2R)\right) + \vec{r} \cdot \vec{J}}
\end{equation}
Now, we do the integrals over $\lambda_1$ and $\lambda_2$ explicitly by using the following representation of a delta function
\begin{equation}
\int {\mathcal D} \xi e^{i C \xi} = \delta[C]
\end{equation}
This is just a functional version of the 1d integral
\begin{equation}
\int_{-\infty}^{\infty} {\rm d} x e^{i k x} = \delta(k)
\end{equation}
Anyway, using this identify, the path integral becomes
\begin{equation}
Z[J] = \int \mathcal{D} \vec{r} \delta\left[|\vec{r}|-R\right] \delta\left[|\vec{r}|-2R\right] e^{i \left(\frac{m}{2} \dot{\vec r}^2  + \vec{r} \cdot \vec{J} \right)}
\end{equation}
Since there is no value of $\vec{r}$ that will make the arguments of both delta functionals zero simultanously, the partition function is identically zero, $Z[J]=0$. Therefore, all transition amplitudes are zero. The theory cannot be unitary, since probabilities have to sum to one but all transition amplitudes are zero, so this is not a sensible quantum theory.
A: It is probably easy to find a Lagrangian producing inconsistent equations of motion. To this end, one can, e.g., start with a complex Lagrangian. It is not obvious that such Lagrangians are necessarily unsatisfactory, because, while they do not seem to make sense in classical mechanics, the relevant path integral in quantum mechanics may be meaningful. Of course, I cannot be sure that such Lagrangians can be useful.
