Showing that a bilinear variation is Lorentz invariant Let $\psi, \chi$ be a spinor (say Dirac). Then the infinitesimal Lorentz variation is given by
$$\delta \psi = -\frac{1}{4}\lambda^{\mu \nu} \gamma_{\mu \nu}\psi$$
then I think that the conjugate is given by
$$ \delta {\psi}^{\dagger} = -\frac{1}{4}\lambda^{\mu \nu} {\psi}^{\dagger} \gamma_{\mu \nu}^{\dagger}.$$ Then we know that  $\bar{\psi}=i\psi^{\dagger}\gamma^0$. My question is how can I show that 
$$\delta( \bar{\chi} \psi) = \delta \bar{\chi} \, \psi + \bar{\chi} \, \delta \psi=0 \,\, ?$$
The second term is ok since it is given by the formula above. The left term though confuses me since 
$$ \delta \bar{\chi} = i\delta \chi^{\dagger} \, \gamma^0 =  -i\frac{1}{4}\lambda^{\mu \nu} {\chi}^{\dagger}\gamma_{\mu \nu}^{\dagger} \, \gamma^0 $$
but I am not sure how to continue from now on. I think I need to use the dagger in the gamma matrix to change the indices and also change the indices of $\lambda^{\mu \nu}$ to get a minus sign. If I do this though I get
 $$ \delta \bar{\chi} =  i\frac{1}{4}\lambda^{\mu \nu} {\chi}^{\dagger}\gamma_{\mu \nu}^{*} \, \gamma^0. $$ Now I do not see how to incorporate this complex conjugate $\gamma_{\mu \nu}$. I think that $\gamma^0$ commutes with $\gamma_{\mu \nu}$ therefore I am left with 
$$ \delta \bar{\chi} =\frac{1}{4}\lambda^{\mu \nu} \bar{\chi}\gamma_{\mu \nu}^{*} $$ but this is not the same as
 $$ \delta \bar{\chi} =\frac{1}{4}\lambda^{\mu \nu} \bar{\chi}\gamma_{\mu \nu} $$
 A: What's wrong with your reasoning is that it is not true that $\gamma^0$ commutes with $\gamma^{\mu\nu}$, as you can see rembering that
$$ \mu \neq \nu \Longrightarrow \gamma^{\mu\nu} = \gamma^\mu \gamma^\nu,$$
hence, for example,
$$ \gamma^{01} \gamma^0 = - \gamma^0 \gamma^{01}.$$

Note: in the following I will use metric signature $(+---)$ and a different convention from the transformation rule, this of course changes nothing in the idea.
Here is a way to do the calculation similar to yours.
From
$$ \tag{1} \delta \psi = \frac{1}{4} \omega_{\mu\nu} \gamma^{\mu\nu} \psi $$
you see that
$$ \tag{2} \delta \bar \psi = \frac{1}{4} \omega_{\mu\nu} \bar \psi \gamma^0 (\gamma^{\mu\nu})^\dagger \gamma^0.$$
You then calculate (remembering that $\gamma^0 \gamma^0 = 1$)
$$ \tag{3} \bar \psi \delta \psi = \frac{1}{4} \omega_{\mu\nu} \bar \psi \gamma^{\mu\nu} \psi,$$
$$ \tag{4} \delta \bar \psi \psi = \frac{1}{4} \omega_{\mu\nu} \bar \psi \gamma^0 (\gamma^{\mu\nu})^\dagger \gamma^0 \psi,$$
$$ \tag{5} \Longrightarrow \delta (\bar \psi \psi) = \frac{1}{4} \omega_{\mu\nu} \bar \psi (\gamma^{\mu\nu}+\gamma^0 (\gamma^{\mu\nu})^\dagger \gamma^0) \psi. $$
The conclusion now follows from the following identity:
$$ \tag{XXX} \gamma^0 (\gamma^{\mu\nu})^\dagger \gamma^0 = -\gamma^{\mu\nu}.$$

I'll give you two ways to prove (XXX):


*

*Start from
$$ \tag{B} \gamma^\mu \gamma^\nu = \eta^{\mu\nu} + \gamma^{\mu\nu},$$
where
$$ \tag{C} \eta^{\mu\nu} = \frac{1}{2} \{ \gamma^\mu,\gamma^\nu \}, $$
$$ \tag{D} \gamma^{\mu\nu} \equiv \frac{1}{2} [\gamma^\mu,\gamma^\nu]. $$
Isolating $\gamma^{\mu\nu}$ in (B) and taking the hermitian conjugate we have
$$ \tag{E} (\gamma^{\mu\nu})^\dagger = (\gamma^\nu)^\dagger (\gamma^\mu)^\dagger - \eta^{\mu\nu}, $$
hence using
$$ \tag{F} (\gamma^\mu)^\dagger = \gamma^0 \gamma^\mu \gamma^0
\Longleftrightarrow \gamma^\mu = \gamma^0 (\gamma^\mu)^\dagger \gamma^0$$
we have
$$ \tag{G} \gamma^0 (\gamma^{\mu\nu})^\dagger \gamma^0
= \gamma^0 (\gamma^\nu)^\dagger \gamma^0 \gamma^0 (\gamma^\mu)^\dagger \gamma^0
- \eta^{\mu\nu}
$$
which using (F) becomes
$$ \tag{H} \gamma^0 (\gamma^{\mu\nu})^\dagger \gamma^0 = \gamma^{\nu\mu} = - \gamma^{\mu\nu}.$$

*Start from (D) and take the hermitian conjugate
$$ \tag{I} (\gamma^{\mu \nu})^\dagger = \frac{1}{2} ( (\gamma^\nu)^\dagger (\gamma^\mu)^\dagger - (\gamma^\mu)^\dagger (\gamma^\nu)^\dagger ), $$
now use (F) to obtain
$$ \tag{L} \gamma^0 (\gamma^{\mu \nu})^\dagger \gamma^0 = \frac{1}{2} ( \gamma^\nu \gamma^\mu - \gamma^\mu \gamma^\nu) = \frac{1}{2} [ \gamma^\nu,\gamma^\mu] = \gamma^{\nu\mu} = -\gamma^{\mu\nu}.$$
A: I'll show you a different approach to the problem first. 
We first define the antisymmetric tensor 
$$\sigma^{\mu\nu}=\frac{i}{2}[\gamma^\mu,\gamma^\nu]$$
A Lorentz transformation $\Lambda$ is defined by three boost and three rotation parameters, i.e. an antisymmetric tensor $\omega_{\mu\nu}$. By looking at how rotations act on the spinor, we conclude that the Lorentz transformation is
$$\psi'(x')=S(\Lambda)\psi(x),\quad S(\Lambda)=\exp(-\tfrac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu})$$
The Clifford algebra tells us that $(\gamma^0)^2=1$ and $(\gamma^i)^2=-1$. Thus, as you may, verify, $\gamma^0$ is hermitean while $\gamma^i$ is antihermitean. This can be expressed as
$$(\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0$$
We define the Dirac adjoint as $\bar\psi=\psi^\dagger\gamma^0$. It follows that $(\sigma^{\mu\nu})^\dagger=\gamma^0\sigma^{\mu\nu}\gamma^0$. Hence $S(\Lambda)^\dagger=\gamma^0\exp(\tfrac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu})\gamma^0$. This leads to
$$\bar\psi'(x')=\psi(x)^\dagger S(\Lambda)^\dagger\gamma^0=\bar\psi(x)\exp(\tfrac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu})$$
Finally
$$\bar\phi'(x')\psi'(x')=\bar\phi(x)\exp(\tfrac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu})\exp(-\tfrac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu})\psi(x)=\bar\phi(x)\psi(x)$$
What went wrong with your calculation is that your $\delta$ does not leave $\gamma^0$ invariant. The gamma matrices have transformation properties of their own under the Lorentz group. I am, however, unsure how to fix your calculation. It seems like it would be very complicated. 
