Bounds for $S=\sqrt{x(y+1)}+\sqrt{y(z+1)}+\sqrt{z(x+1)}$ Let $x,y,z\in[0,2]$ and $x+y+z=3$. What are the maximum and minimum of $$S=\sqrt{x(y+1)}+\sqrt{y(z+1)}+\sqrt{z(x+1)}?$$
When $x=y=z=1$, we have $S=3\sqrt{2}$.
For the upper bound, we can use the AM-GM inequality to get $$S\leq\frac{1}{2}(x+y+1+y+z+1+z+x+1)=\frac{9}{2},$$ but equality cannot hold since we need $x=y+1$, etc. simultaneously, which cannot happen.
 A: A little numerical investigation gives that the maximum is attained for $x=y=z=1$ and the minimum for $x=1,y=2$ and $z=0$. An intuitive proof of this fact can be achieved by noticing that $F(x,y,z) = \sum_{cyc} \sqrt{x(y+1)}$ is concave in every variable and thus, the minimum should be attained at extremal points and the maximum in the interior of the domain.
To prove rigorously the upper bound apply Cauchy Schwarz inequality to see that
$$ S^2 \leq (x+y+z)(y+1+z+1+x+1) = 18.$$
This implies that $S \leq 3\sqrt{2}$ and thus the maximum is $3\sqrt{2}$ and is attained only for $x=y=z=1$.
In order to find the minimal value we could suppose that $x \leq y \leq z$ and we look what happens if we replace the three values by $x-\varepsilon\leq y \leq z+\varepsilon$. We define
$$ f(\varepsilon)=\sqrt{(x-\varepsilon)(y+1)}+\sqrt{y(z+\varepsilon+1)}$$
and we note that
$$ f'(\varepsilon) = -\sqrt{\frac{y+1}{x-\varepsilon}}+\sqrt{\frac{y}{z+\varepsilon+1}} $$
Note that since $\frac{y+1}{x}>\frac{y}{z+1}$ (simple consequence of the ordering) for small $\varepsilon$ the derivative is negative. Thus $f(\varepsilon)<f(0)$. This means that increasing the distance between $x$ and $z$ decreases the value of the function.
We can choose $\varepsilon$ as large as the domain allows it until we reach an extremal case: $x=0$ or $z=2$.
Case  1. $x=0$. Then $y+z=3$ and we want to minimize $g(y,z) = \sqrt{y(z+1)}+\sqrt{z}$. This is a one dimensional problem. It is not hard to see that the same principle applies. If we enlarge the distance between $y$ and $z$ then the value of the function decreases. Thus the minimum is attained for $y=1,z=2$.
Case 2. $z=2$. The $x+y=1$ and we want to minimize $g(x,y) = \sqrt{x(y+1)}+\sqrt{3y}+\sqrt{2(x+1)}$. The same type of analysis applies and we get that the minimum is found for $x=0,y=1$.
Thus the minimal value is equal to $3$ and is attained for $x=0,y=1,z=2$.
