Prove $\sqrt{a + ab} + \sqrt{b} + \sqrt{c} \ge 3$ for $c = \min(a, b, c)$ and $ab + bc + ca = 2$ 
Problem 1: Let $a, b, c \ge 0$ with $c = \min(a, b, c)$ and $ab + bc + ca = 2$. Prove that
$\sqrt{a + ab} + \sqrt{b} + \sqrt{c} \ge 3$.

Background: This problem was proposed by csav10@AoPS
(https://artofproblemsolving.com/community/q1h2321937p18543869)
which is a variant of the following:

Problem 2: Let $a, b, c \ge 0$ with $ab + bc + ca = 2$. Prove that $\sqrt{a+ab} + \sqrt{b+bc} + \sqrt{c+ca} \ge 3$.

(Proposed by KaiRain@AoPS. See https://artofproblemsolving.com/community/c6h1905412.
Also see: https://artofproblemsolving.com/community/q1h2308247p18537783.)
For Problem 2, yleo@AoPS and csav10@AoPS gave nice solutions, respectively.
Also, I solved Problem 2 by the Buffalo Way (BW) which is a computer solution.
For Problem 1, I have not yet solved it by the Buffalo Way (BW).
By the way, Equality case: $(a, b, c) = (2, 1, 0)$.
Any comments and solutions are welcome and appreciated.
Add the Buffalo Way solution (outline) for Problem 2
Fact 1: If $b \ge a \ge c \ge 0$, then
$$\sqrt{a + ab} + \sqrt{b + bc} + \sqrt{c + ca} \ge \sqrt{b + ba} + \sqrt{a + ac} + \sqrt{c + cb}.$$
From Fact 1, we assume that $a \ge b \ge c$.
After homogenization, we need to prove that
$$\sqrt{\frac{a \sqrt{\frac{ab+bc+ca}{2}} + ab}{\frac{ab+bc+ca}{2}}} +
\sqrt{\frac{b \sqrt{\frac{ab+bc+ca}{2}} + bc}{\frac{ab+bc+ca}{2}}} +
\sqrt{\frac{c \sqrt{\frac{ab+bc+ca}{2}} + ca}{\frac{ab+bc+ca}{2}}} \ge 3.$$
Denote
$$X = \frac{1}{4}\cdot\frac{a \sqrt{\frac{ab+bc+ca}{2}} + ab}{\frac{ab+bc+ca}{2}},
Y = \frac{b \sqrt{\frac{ab+bc+ca}{2}} + bc}{\frac{ab+bc+ca}{2}},
Z = \frac{c \sqrt{\frac{ab+bc+ca}{2}} + ca}{\frac{ab+bc+ca}{2}}.$$
We need to prove that $2\sqrt{X} + \sqrt{Y} + \sqrt{Z} \ge 3$.
By using $\sqrt{u} \ge \frac{2u}{1+u}$ for $u\ge 0$, it suffices to prove that
$$\frac{4X}{1+X} + \frac{2Y}{1+Y} + \frac{2Z}{1+Z} \ge 3.$$
After some manipulation, it suffices to prove $A\sqrt{\frac{ab+bc+ca}{2}} + B \ge 0$ where $A, B$ are some polynomials and $A \ge 0$.
It suffices to prove that $A^2\cdot \frac{ab+bc+ca}{2} \ge B^2$.
BW works.
 A: Partial answer
Let $f = \sqrt{a + ab} + \sqrt{b} + \sqrt{c} - 3$ and $g = ab + bc + ca -2$, so we are to show $f\ge 0$ under the condition $g =0$.
Start by noting that if $a=b=c=c_0$, we have  $c_0 = \sqrt{2/3} \simeq 0.8165$ and $f = 0.0251$ so the condition is very tight.
Nevertheless, in this partial answer the inequality can be proved for small $c$, and for $c$ in the (larger) vicinity of $c_0$. Indications are given to extend these calculations.
Treat $c$ as a parameter. First, note that for $a\ge b$, we observe for the exchange of $a\leftrightarrow b$ that  $\sqrt{a + ab} + \sqrt{b} + \sqrt{c} - 3 \le \sqrt{b + ab} + \sqrt{a} + \sqrt{c} - 3$, hence we need to consider the case $a\ge b$ only.
Since $c$ is required to be the smallest variable, we consider $a\ge b \ge c$.
Case 1: small $c$.
We show that the equality condition $(a,b,c) = (2,1,0)$ is actually a minimum of $f$. A convenient way to do so is variational calculus, where $c$ is arbitrary but fixed. This gives
$
0 = dg = (b+c)da + (a+c)db$
and
$ 2 \sqrt{a + ab}\sqrt{b} \; df = \sqrt{b} (1+b) \; da + (a \sqrt b + \sqrt{a + ab}) \; db$
For the RHS to become zero we insert the $dg=0$-equation and get
$  d \tilde f = [\sqrt{b} (1+b) (a+c) -   (a \sqrt b + \sqrt{a + ab}) (b+c)] \, da  $.
For $c=0$, $  d \tilde f =  0$ gives
$   \sqrt{a}=   \sqrt{1 + b} \sqrt{b}$. Now inserting, from $g=0$, the condition $a = 2/b$, gives $2 = b^2 + b^3$ which has the only positive solution $b=1$, from which follows again $a = 2$.  This establishes that $(a,b,c) = (2,1,0)$ is actually a minimum of $f$.
Now in the vicinity of that minimum, i.e. for small $c > 0$, the $f$-equation itself guarantees the inequality. The reason is that $  d \tilde f = 0$ tells us that the value of $f$  will shift away in linear dependence with $c$, as $a$ and $b$ vary to obey $  d \tilde f = 0$.  In contrast, in $f$ the additional term $\sqrt c$ increases $f$, where this term has an infinite growth rate at $c=0$ (illustrated also in  the second figure below).  Hence for small  $c > 0$, $f$ will always move in positive direction with $c$.
This behavior can be utilized with the following sketch: use $d \tilde f = 0$ and $g=0$ to calculate the (small) change of $a$ and $b$ once $c$ moves away from $0$. With this change, calculate the new $f$ where the first terms change linearly and the last term goes with $\sqrt c$.  Then, equate the "worst" change in the first terms with  $\sqrt c$ to get an estimate for the highest $c$  where $f \ge 0$ holds.
Case 2: large $c$.
The largest $c$ that can be attained is  $c_0 = \sqrt{2/3}$, as $c$ must be the smallest variable.  In this case,  $a=b=c=c_0$, $f = 0.0251$, and variations of $(a,b,c)$ under the condition $g=0$ change $f$ linearly with $c$. Hence, in some environment $f$ is guaranteed to stay positive.
Indeed, here we can estimate $f$. Method 1: As $a$ is the largest of the variables, the smallest $a$, for given $c$, that can be attained, is observed when $a=b$. Under $g=0$, this gives $a = \sqrt{2 + c^2} - c$. The smallest $b$ is $b=c$. For $a=b=c=c_0$ both estimates are tight, so for some range of  $c < c_0$, we can ensure  $f\ge 0 $ to hold. We have
$f \ge  \hat f = \sqrt{(\sqrt{2 + c^2} - c)} \, \sqrt{1 + c} + 2\sqrt{c}  - 3$ and this holds until $\hat f = 0$ which is attained for $c_{min}  \simeq 0.793$.
Method 2: Inserting $a$ from $g =0$ gives $f = \sqrt{2-b c} \, \frac{\sqrt{1 + b}}{\sqrt{c + b}} + \sqrt{b} + \sqrt{c} - 3$. Now  $\frac{\sqrt{1 + b}}{\sqrt{c + b}} \ge \frac{\sqrt{1 + b_{max}}}{\sqrt{c_{max} + b_{max}}}$ and, by AM-GM, $\sqrt{b} + \sqrt{c} \ge 2 \sqrt[4]{b c}$. Letting $bc = x$ this gives   $$f \ge \hat f =  \sqrt{2-x} \, \frac{\sqrt{1 + b_{max}}}{\sqrt{c_{max} + b_{max}}} + 2 \sqrt[4]{x} - 3$$ which is a function of $x$ only, with $b_{max} = \max\{\sqrt{2 + c^2} - c\} = \sqrt{2}$ and $c_{max} = c_0$. Again, all estimates are tight for $a=b=c=c_0$, which entails $x = \frac23$ and $f = 0.0251$.  So $\hat f = 0$ can be used to calculate the smallest $x$ where the estimate holds, which is given for $\hat x \simeq 0.632$. Since $c = \hat x/b$, this makes it feasible to reduce $c$ while increasing $b$, however this is only possible until $b$ reaches its highest value. For given $c$, this value is  $\sqrt{2 + c^2} - c$ which gives $c(\sqrt{2 + c^2} - c) = 0.632$ or $c_{min} =  0.737$, which is better than the bound with method 1.
Again, this can be improved, if better values for $b_{max}$ and $c_{max}$  are found. For example, with the result just obtained, one can iterate the method and reduce $c$ once again, where now it is established that $c_{max} = 0.737$ can be used. Also, the estimate $b_{max} = \max\{\sqrt{2 + c^2} - c\}$ needs the smallest value of $c$ which goes into consideration, where $c=0$ was the most conservative one. Here, once, from the extension of case 1, a highest value of $c$  is known for which the inequality holds, this value can be used.
Case 3: "small $c$" $ < c < 0.737$.
To be done. Indications are given above.
Here is a figure which shows what to expect, and consequently what to do:

In green is the location of the minimum, i.e. the solution of $df = 0$ under $g=0$. In red is the range for $b$, i.e. $c \le b \le \sqrt{2 + c^2} \, - c$. For  $c=0$, the minimum is to be found at $b_{min}=1$, as was calculated above. As $c$ increases, $b_{min}$ gets smaller. For $c \simeq 0.345$, the location of $b_{min}$ reaches the lower limit  of the range for $b$. Hence for the remaining $ 0.345 < c < \sqrt{2/3}$, the smallest $f$ is to be found at $b_{min} = c$. This makes it easy to show the inequality in this interval of $c$: we have
$f \ge f_{min}^* = \sqrt{2- c^2} \, \frac{\sqrt{1 + c}}{\sqrt{2 c}} + 2\sqrt{c} - 3$ and this is clearly positive, see the following plot (created with values from numeric evaluations). In that plot, $f_{min}$ is plotted over the first $c$-interval $\mbox{[0 $\,$ 0.345]}$ where the value of $b$ is taken as the free minimum, from the solution of $df = 0$ under $g=0$ (blue section). The green circle is there to mark the transition to the second $c$-interval [$0.345 \,$ $\sqrt{2/3}$] where $f_{min}^*$ is plotted (see above) which is  taken at $b = c$ (orange section). Note the rise of $f_{min}$ with infinite rate at $c=0$ acording to the $\sqrt{c}$-term in $f$, as discussed above under case 1: small $c$.

Hence the agenda for analytic work is:

*

*establish for $c < 0.345$ that $f$ has a free minimum w.r.t. $b$, and that at that minimum, $f_{min}(c) > f_{min}(c=0) =0$.

*establish for $c > 0.345$ that $f(b,c) \ge f(c,c)$, and then  $f(c,c) > 0$ can be shown easily (see above), which gives the desired result.

