<s>[INST] 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?
 [/INST] 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 $1+\sqrt{\frac{55}{6}}$.
</s>