If $a+b+c+d=0$ and $\{a,b,c,d\}\subset[-1,1]$ so $\sum\limits_{cyc}\sqrt{1+a+b^2}\geq4$

Question

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.

Answer 1

Some thoughts

(For 3-variable problem, see: Prove $\sum_{\mathrm{cyc}} \sqrt{34x^2 + 28y^2 + 7z^2 - xy - 28yz + 41zx} \ge 9x + 9y + 9z$)

It suffices to prove the following:

Problem 1: Let $a, b, c, d \ge -1$ with $a+b+c+d = 0$. Prove that $$\sqrt{1 + a + b^2} + \sqrt{1 + b + c^2} + \sqrt{1 + c + d^2} + \sqrt{1 + d + a^2} \ge 4.$$ (Thank @tthnew for verifying it by his routine.)

Problem 1 is equivalent to Problem 2 and Problem 3 below.

Problem 2: Let $x, y, z, w \ge 0$ with $x + y + z + w = 4$. Prove that $$\sqrt{x + (y-1)^2} + \sqrt{y + (z-1)^2} + \sqrt{z + (w-1)^2} + \sqrt{w + (x-1)^2} \ge 4.$$

Problem 3: Let $x, y, z, w \ge 0$. Prove that $$\sum_{\mathrm{cyc}} \sqrt{5x^2 + 9y^2 + z^2 + w^2 - 2xy + 6xz + 6xw - 6yz - 6yw + 2zw} \ge 4(x+y+z+w).$$

Answer 2

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.