Is there a generalization of Araki-Lieb-Thirring inequality for four matrices? It is known that $$ Tr[(AB)^n] \leq Tr (A^n B^n)$$ for $A$, $B$ positive semi-definite matrices. I am looking of a generalization of it for a product of $4$ positive semi-definite matrices of the the form, 
$$Tr [(ABCD)^n] \leq Tr [(AB)^{2n} ] Tr [(DC)^{2n}]$$ 
Is there any such inequality? Can someone suggest an approach if the above thing can be proved or disproved? 

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*The inequality you propose seems not very homogeneous ... in your case, just take $A=B=C=D$ and it becomes
$$\mathrm{Tr}(A^{4n}) ≤ \mathrm{Tr}(A^{4n})^2,$$
which is obviously false for general matrices.

*The homogeneous version could be
$$\mathrm{Tr}((ABCD)^{n}) ≤ \mathrm{Tr}((AB)^{2n})^\frac{1}{2}\, \mathrm{Tr}((CD)^{2n})^\frac{1}{2}.$$
But it seems to also be false in the general case. To prove that, take for example $CD = BA$ and you will obtain
$$\mathrm{Tr}(|BA|^{2n}) ≤ \mathrm{Tr}((AB)^{2n})^\frac{1}{2}\, \mathrm{Tr}((BA)^{2n})^\frac{1}{2}$$
where I used the notation for the absolute value value of a matrix defined by $|A|^2 = A^* A$, which gives $|BA|^2 = AB^2A$. By using the cyclicity of the trace, it is easy to see that $\mathrm{Tr}((BA)^{2n}) = \mathrm{Tr}((AB)^{2n})$, so your inequality becomes
$$\mathrm{Tr}(|BA|^{2n}) ≤ \mathrm{Tr}((BA)^{2n}),$$
which is false in general.

*An inequality in the idea of what you are looking for could be
$$\mathrm{Tr}((ABCD)^{n}) ≤ \mathrm{Tr}(|AB|^{2n})^\frac{1}{2}\, \mathrm{Tr}(|CD|^{2n})^\frac{1}{2},$$
which follows from Hölder's inequality for matrices. It can equivalently be written
$\mathrm{Tr}((ABCD)^{n}) ≤ \mathrm{Tr}((A^2B^2)^{n})^\frac{1}{2}\, \mathrm{Tr}((C^2D^2)^{n})^\frac{1}{2}$
using the definition of the absolute value and the cyclicity of the trace. The Araki-Lieb-Thirring inequality then tells you that you can also obtain
$$\mathrm{Tr}((ABCD)^{n}) ≤ \mathrm{Tr}(A^{2n}B^{2n})^\frac{1}{2}\, \mathrm{Tr}(C^{2n}D^{2n})^\frac{1}{2}.$$
