Find the minimum of $a^4+b^4+c^4+d^4+a^2+b^2+c^2+d^2$. 
Let $a,b,c,d$ be real numbers such that $a+b+c+d=0$ and $abcd=1$. Find
the minimum value of $a^4+b^4+c^4+d^4+a^2+b^2+c^2+d^2$.

By $\text{Vieta}$'s theorem, $a,b,c,d$ are the roots of the equation $x^4+sx^2+tx+1=0$. Thus
\begin{align*}
&~~~~~~a^4+b^4+c^4+d^4+a^2+b^2+c^2+d^2\\&=(1-s)(a^2+b^2+c^2+d^2)-4\\
&=(1-s)[(a+b+c+d)^2-2(ab+ac+ad+bc+bd+cd)]-4\\
&=(1-s)(0-2s)-4=2(s^2-2s-4).
\end{align*}
Now, we only need find the range of $s$. But how to do?
 A: Let $f(a,b,c,d,\lambda,\mu)=\sum\limits_{cyc}(a^4+a^2)+\lambda(a+b+c+d)+\mu(abcd-1).$
Thus, in the minimal point we need $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=\frac{\partial f}{\partial d}=\frac{\partial f}{\partial \lambda}=\frac{\partial f}{\partial a}=\frac{\partial f}{\partial \mu}=0.$$
Let $a$, $b$ and $c$ be different numbers.
Thus, since $$4a^3+2a+\lambda+\mu bcd=0$$ or
$$4a^4+2a^2+\lambda a+\mu abcd=0,$$ which with $$4b^4+2b^2+\lambda b+\mu abcd=0$$ gives
$$4a^4+2a^2+\lambda a=4b^4+2b^2+\lambda b$$ or
$$(a-b)(4(a^2+b^2)(a+b)+2(a+b)+\lambda)=0$$ or
$$4(a^2+b^2)(a+b)+2(a+b)+\lambda=0.$$
Similarly $$4(a^2+c^2)(a+c)+2(a+c)+\lambda=0$$ and $$4(c^2+b^2)(c+b)+2(c+b)+\lambda=0,$$ which gives
$$4(a^2+b^2)(a+b)+2(a+b)=4(a^2+c^2)(a+c)+2(a+c)$$ or
$$(b-c)\left(1+2\sum_{cyc}(a^2+ab)\right)=0$$ or
$$1+\sum_{cyc}(a+b)^2=0,$$ which is impossible.
We got a contradiction.
Thus, for any three variables at least two of them are equal.
Id est, it's enough to find a minimal value in the following cases.

*

*Three  variables are equal.

Let $b=c=a$.
Thus, $d=-3a$ and $-3a^4=1,$ which is impossible;


*$a=b$ and $c=d$.

In this case $a+c=0$ and $a^2c^2=1,$ which gives $(a,b,c,d)=(1,1,-1,-1)$.
In the general we obtain any symmetric permutations of this, which gives a value $8$
and it's a minimal value because the maximal value does not exist.
Also, by AM-GM $$\sum_{cyc}(a^4+a^2)\geq4|abcd|+4\sqrt{|abcd|}=8,$$ and the equality occurs.
