If $abcd=1$ then $a^4b+b^4c+c^4d+d^4a\geq a+b+c+d$ 
I want to prove that for all $a>0$, $b>0$, $c>0$ and $d>0$ with $abcd=1$:  $$a^4b+b^4c+c^4d+d^4a\geq a+b+c+d.$$

I think that this should be provable by AM-GM inequality, but I could not manage to prove it. Can you give me a hint?
Best wishes
 A: The hint:
Write 
$$xa^4b+yb^4c+zc^4d+td^4a\geq a^2bcd,$$
where $x$, $y$, $z$ and $t$ are non-negatives such that $x+y+z+t=1$ 
and after using AM-GM solve the system, which you got.
I got $x=\frac{23}{51}$, $y=\frac{7}{51}$, $z=\frac{11}{51}$ and $t=\frac{10}{51}$,
which gives the following magical solution in one line.
$$\sum_{cyc}a^4b=\frac{1}{51}\sum_{cyc}(23a^4b+7b^4c+11c^4d+10d^4a)\geq$$
$$\geq\frac{1}{51}\sum_{cyc}\left(51\sqrt[51]{a^{23\cdot4+10}b^{23+7\cdot4}c^{7+11\cdot4}d^{11+10\cdot4}}\right)=\sum_{cyc}a^2bcd=\sum_{cyc}a.$$
Also we can use C-S here.
C-S inequality it's the following.
Let $b_i>0$. Prove that:
$$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+...+\frac{a_n^2}{b_n}\geq\frac{(a_1+a_2+...+a_n)^2}{b_1+b_2+...+b_n}.$$
Hence, by C-S we obtain:
$$\sum_{cyc}a^4b=\sum_{cyc}\frac{a^3}{cd}=\sum_{cyc}\frac{(a^2)^2}{acd}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}a^2\right)^2}{\sum\limits_{cyc}acd}=\frac{(a^2+b^2+c^2+d^2)^2}{abc+abd+acd+bcd}.$$
Thus, it remains to prove that
$$(a^2+b^2+c^2+d^2)^2\geq(a+b+c+d)(abc+abd+acd+bcd),$$ which is obvious by Muirhead
because $(4,0,0,0)\succ(2,1,1,0)$, $(2,2,0,0)\succ(2,1,1,0)$ and $(2,2,0,0)\succ(1,1,1,1)$.
