How prove this ineuality$\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1}\le\sqrt{xy+yz+zx+9}$

Question

let $$x,y,z\in(-1,1), x+y+z=-xyz$$

show that $$\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1}\le\sqrt{xy+yz+zx+9}$$

This problem is my frends ask me,I remenber this is old inequality,But Now I can't it

$$x+y+z+3+2\sum_{cyc}\sqrt{xy+x+y+1}\le xy+yz+xz+9$$

Answer 1

Do as you said and we only have to prove $$2\sum_{cyc}\sqrt{xy+x+y+1}\le xy+yz+xz+6-x-y-z$$

This is right, since $\displaystyle x+y+z=-xyz$ and $x,y,z,xy,yz,xz\in(-1;1)$, we have:

$2\sqrt{xy+x+y+1} =2\sqrt{xy+x+y+1-x-y-z-xyz} =2\sqrt{xy-xyz+1-z} =2\sqrt{(1-z)(1+xy)} \le 1-z+1+xy$

Do this with the other two and sum up, we have the inequality.