If $a+b+c+d=0$ and $\{a,b,c,d\}\subset[-1,1]$ so $\sum\limits_{cyc}\sqrt{1+a+b^2}\geq4$ Let $\{a,b,c,d\}\subset[-1,1]$ such that $a+b+c+d=0$. Prove that:
$$\sqrt{1+a+b^2}+\sqrt{1+b+c^2}+\sqrt{1+c+d^2}+\sqrt{1+d+a^2}\geq4$$
I tried Holder and more, but without success. 

Let $ {x,y,z}\in[-1, 1]$,and $x+y+z=0 $
Prove: $$\sqrt{1+x+y^2}+\sqrt{1+y+z^2}+\sqrt{1+z+x^2}\geq 3$$
Proof:
First we will show:
If $ab\geq 0, $ then:$\sqrt{1+a}+\sqrt{1+a}\geq 1+\sqrt{1+a+b}$
This is obvious！
Note that at least two of $ x+y^2, y+z^2 $ and $ z+x^2 $ have the same positive and negative values。Without loss of generality, we assume:$(x+y^2)(y+z^2)\geq 0$ then we have:$$\sqrt{1+x+y^2}+\sqrt{1+y+z^2}+\sqrt{1+z+x^2}$$ $$\geq 1+\sqrt{1+x+y^2+y+z^2}+\sqrt{1+z+x^2}$$ $$=1+\sqrt{(\sqrt{1-z+z^2})^2+y^2}+\sqrt{(\sqrt{1+z})^2+x^2}$$ $$\geq 1+\sqrt{(\sqrt{1-z+z^2}+\sqrt{1+z})^2+(x+y)^2}$$ $$ =1+\sqrt{(\sqrt{1-z+z^2}+\sqrt{1+z})^2+z^2}$$ 
So, just prove:$$(\sqrt{1-z+z^2}+\sqrt{1+z})^2+z^2\geq 4$$ $$\Longleftrightarrow 2z^2+2\sqrt{1+z^3}\geq 2$$ $$\Longleftrightarrow z^2(2-z)(z+1)\geq 0$$
This is clearly true because $|z|\leq 1$.Equal sign is established if and only if $x=y=z=0$
That's all,I hope it's works for you.
PS:Even if this method can be used, but the calculation is too complex that it is almost impossible to complete. The case of three variables is a good example, so I will not withdraw the answer.
