Find the minimum value of $\sqrt{a+1}+\sqrt{2b+1}+\sqrt{3c+1}$ if $a+b+c=1$ 
Let $a,b,c\ge 0$, and $a+b+c=1$. Find the minimum value of
$$\sqrt{a+1}+\sqrt{2b+1}+\sqrt{3c+1}$$

I think the minimum value is $\sqrt{2} + 2$? when $a=1,b=c=0$.
Of course, I can't prove it. Can anyone help?
 A: The objective function is concave in each of its variables, and the domain is closed and convex. so the minimum will occur when the variables take extreme values allowed. It is easy to see that with the constraint, this is possible when two among $a,b,c$ is zero, and the minimum is $2+\sqrt2$ obtained when $a=1, b=c=0$. 
A: Let
$$
U=\{(x,y,z)\in\mathbb{R}^3:x+y+z=1\text{ and }x,y,z\ge 0\}\\
S=\{(x,y,z)\in\mathbb{R}^3:x+y+z=1\text{ and }x,y,z>0\}.
$$
Define $f,g:U\to \mathbb{R}$ such that
\begin{align}
f(a,b,c)&=\sqrt{a+1}+\sqrt{2b+1}+\sqrt{3c+1}\\
g(a,b,c)&=a+b+c-1,
\end{align}
then
\begin{align}
\nabla f&=\left(\frac{1}{2\sqrt{a+1}},\frac{1}{\sqrt{2b+1}},\frac{3}{2\sqrt{3c+1}}\right)\\
\nabla g&=(1,1,1).
\end{align}
If a local minimum exists in $S$, there is $\lambda$ such that $\nabla f = \lambda \nabla g$, Thus
\begin{align}
a&=\frac{1}{4\lambda^2}-1\\
b&=\frac{1}{\lambda^2}-\frac{1}{2}\\
c&=\frac{9}{4\lambda^2}-\frac{1}{3}\\
\end{align}
Substitute $a,b,c$ in $g(a,b,c)=0$, then we get
$$
\frac{1}{\lambda^2}=\frac{17}{9}.
$$
But in this case, we find $a <0$, so there is no critical point of $f$ restricted to $g(x,y,z)=0$ in $S$.
Now we will set $c$ as $0$, then
$$
f(a)=\sqrt{a+1}+\sqrt{3-2a}
$$
and so
$$
f'(a)=\frac{1}{2\sqrt{a+1}}-\frac{1}{\sqrt{3-2a}}.
$$
Since $a\ge 0$,
$$
\frac{1}{2\sqrt{a+1}}-\frac{1}{\sqrt{3-2a}} > 0
$$
and so $f$ increases. Therefore $f(a,b,c)$ has its minimum $2+\sqrt{2}$ at $(1,0,0)$.
Exercise?: $f$ restricted to $g(x,y,z)=0$ has its maximum $1+\sqrt{\frac{55}{6}}$.
A: We have
\begin{align*}
 &\sqrt{a + 1} + \sqrt{2b + 1} + \sqrt{3c + 1} - (\sqrt{2} + 2)\\
 =\, & (\sqrt{a + 1} - \sqrt{2}) + (\sqrt{2b + 1} - 1) + (\sqrt{3c + 1} - 1) \\
 =\, & \frac{a - 1}{\sqrt{a + 1} + \sqrt{2}} + \frac{2b}{\sqrt{2b + 1} + 1} + \frac{3c}{\sqrt{3c + 1} + 1}\\
 =\, & \frac{-b - c}{\sqrt{a + 1} + \sqrt{2}} + \frac{2b}{\sqrt{2b + 1} + 1} + \frac{3c}{\sqrt{3c + 1} + 1}\\
 \ge\, & \frac{-b - c}{1 + \sqrt{2}} + \frac{2b}{\sqrt{3} + 1} + \frac{3c}{\sqrt{4} + 1}\\
 =\, & (\sqrt{3} - \sqrt{2})b + (2 - \sqrt{2})c \\
 \ge\, &0.
\end{align*}
Also, if $a = 1, b = c = 0$, then
$\sqrt{a + 1} + \sqrt{2b + 1} + \sqrt{3c + 1} = \sqrt{2} + 2$.
Thus, the minimum of $\sqrt{a + 1} + \sqrt{2b + 1} + \sqrt{3c + 1}$ is $\sqrt{2} + 2$.
