How to check whether the production function below is neoclassical or not Say I'm given the production function of the form

\begin{equation*}
Y= F(K,L,A) = (\alpha K^{\epsilon} + (1 - \alpha)Y^{\epsilon})^{\frac{1}{\epsilon}}
\end{equation*}

Where $\epsilon, \alpha \in (0,1)$. 

Now I need to check three things:


That marginal products are positive and they are decreasing
Inada conditions are satisfied
Essentiality
Linear homogeneity


The conditions 2 and 4 imply the condition 3 so no need to check for that. When taking derivatives with respect to $K$ I see that

\begin{equation*}
F_K = \alpha  K^{\epsilon -1}(\alpha K^{\epsilon} + (1 - \alpha)Y^{\epsilon})^{\frac{1 - \epsilon}{\epsilon}} &gt;0
\end{equation*}

Which is positive. Now I need to show that $F_{KK}&lt; 0$. See:

\begin{equation*}
F_{KK} = \alpha (\epsilon -1) K^{\epsilon -2}(\alpha K^{\epsilon} + (1 - \alpha)Y^{\epsilon})^{\frac{1 - \epsilon}{\epsilon}} + \alpha^2 K^{2\epsilon -2} \frac{1- \epsilon}{\epsilon} (\alpha K^{\epsilon} + (1- \alpha) L^{\epsilon})^{\frac{1}{\epsilon}-2}
\end{equation*}

However I couldn't find a way to decide on the sign of this expression. Any help would be appreciated. Thanks.
C.E.S production functions are indeed tiresome to differentiate, and it is very easy to make a mistake. But hey, this (algebraic manipulation) is part of the job, and one needs to practice moving around symbols , creating common factors, etc. 

But there is another way to show that $F_{KK} &lt;0$ in totally abstract terms.

First, prove 4. (linear homogeneity), it is easy.  

Second, exploit Euler's Homogeneous Function Theorem, and decompose $Y$ using the first derivatives and the variables (this is a very well known result).

Third, re-arrange the previous to obtain an alternative abstract expression of the first derivative of say, capital.

Fourth, differentiate what you got in step Three, to obtain an expression for $F_{KK}$. 

All the above are done using abstract notation $F_K, F_{KK}$,etc and nowhere the actual C.E.S functional form. After step Four, you will need to calculate the cross-partial, which is easy, even for the C.E.S. function.