<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] 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 $\sqrt{2} + 2$.
</s>