TITLE: Prove $\sqrt{a + ab} + \sqrt{b} + \sqrt{c} \ge 3$ for $c = \min(a, b, c)$ and $ab + bc + ca = 2$
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: Full answer
This is an answer with a different method than my other (partial) answer, therefore a new one.
Outline: this answer establishes a polynomial inequality and then shows that the polynomial will have no zeroes which gives the proof. To do so, we apply a method where all coefficients of this polynomial are inspected, which makes the proof somewhat technical.
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 $a\ge b \ge c$.
As expressed in the problem statement, a single minimum of the target function is expected at $b=1$ and $c=0$, with equality.
To establish a polynomial, first replace all variables by their squares. Then we need to show that
$$
f = a \sqrt{1 + b^2} + b + c - 3 \ge 0
$$
with the condition $ g = a^2b^2 + b^2 c^2 + c^2 a^2 - 2 = 0$. The condition for $f$ can be written
$$
a^2 (1 + b^2) - (b + c - 3)^2 \ge 0
$$
(Note $b + c - 3 <0$ in the case of concern, $a\ge b\ge c$, so this is indeed equivalent to $f \ge 0$.)
From $g$, replacing  $a^2 = \frac{2 - b^2c^2}{b^2 + c^2}$ then gives that we have to prove
$$
F = (2 - b^2c^2) (1 + b^2) - (b^2 + c^2)(b + c - 3)^2 \ge 0
$$
which is now a bivariate polynomial with conditions on the range of $b$ and $c$ as follows (for qualitative illustration see also the first figure in my other answer):
We turn to the condition $c = \min(a, b, c)$.
Due to the condition $g$, we have that $c \le b$ for $b \in [0 \quad \sqrt[4]{\frac23}]$. If $b$ gets larger  than $\sqrt[4]{\frac23}$, we have to ensure $a \ge b$ which entails the limit $a=b$ in $g$, giving $c \le \sqrt{\frac{2 - b^4}{2 b^2}}$ for  $b \in [ \sqrt[4]{\frac23}  \quad \sqrt[4]{2} ]$. The upper limit for $b$ is given when $a=b$ and $a^2 b^2 = 2$, which forces $c=0$ in $g$. We  will come back to these two ranges for $b$ later.
We can now replace $c = x b$ with $x \in (0 \quad 1]$. Doing so we have to show
$$
F = (2 - b^4 x^2) (1 + b^2) - b^2(1 + x^2)(b + b x - 3)^2 \ge 0
$$
Regard $F$ as a polynomial in $x$ with parameter $b$. For some permitted values $(x, b)$ we have $F > 0$. For proving the claim, we establish now that for $x \in (0 \quad 1]$, this polynomial has no roots, hence $F>0$ everywhere in the permitted range of $b$ and $c$. To do so, we use Budan's theorem. The theorem says that if the number of sign changes of the coefficients in $F(x)$ equals the number of sign changes of the coefficients in $F(x+1)$, then the polynomial $F(x)$ has no roots in  $(0 \quad 1]$.
We have
$$
F(x)  = [- b^4 + 6b^3 - 7b^2 + 2] + [6 - 2b] b^3 x + [- b^4 - 3b^2 + 6b - 9] b^2 x^2 +  [6 - 2b] b^3 x^3 +
[-1]b^4 x^4
$$
Regardless of the value of $b \in [0  \quad \sqrt[4]{2} ]$, we have the following signs, symbolically denoted by
$$
F(x)  = ["+"] + ["+"]  x + ["-"] x^2 +  ["+"] x^3 +
["-"] x^4
$$
which makes three sign changes.
For $b \in [0 \quad 1]$, we have
$$
F(x+1)  = [- b^6 - 9 b^4 + 24 b^3 - 16 b^2 + 2] + [- 2 b^4 - 18 b^2 + 36 b - 18] b^2 x + [- b^4 - 15 b^2 + 24 b - 9] b^2 x^2 +  6 [1 - b] b^3 x^3 +
[-1]b^4 x^4
$$
Again we have the following signs, symbolically denoted by
$$
F(x+1)  = ["+"] + ["-"]  x + ["-|+"] x^2 +  ["+"] x^3 +
["-"] x^4
$$
where $["-|+"]$ indicates that, depending on the value of $b \in [0 \quad 1]$, this coefficient has different signs. No matter what this sign, however, there are three sign changes in the coefficients of $F(x+1)$.
For $b \in [1 \quad \sqrt[4]{2} ]$, we have   $c \le \sqrt{\frac{2 - b^4}{2 b^2}}$. To make matters easier, we increase the range of $c$ by the wider limit $c \le 3 - 2 b$. We use this limit to introduce $c = x( 3 - 2 b)$ with  $x \in [0 \quad 1]$.
Then we have
$$
F (x) = (2 - x^2 b^2(3 - 2 b)^2) (1 + b^2) - (b^2 + x^2(3 - 2 b)^2)(b + x(3 - 2 b) - 3)^2
$$
This gives
$$
F(x)  = [- b^4 + 6 b^3 - 7 b^2 + 2] +[4 b^2 - 18 b + 18] b^2 x + [- 4b^6 + 12 b^5 - 21 b^4 + 60 b^3 - 135 b^2 + 162 b - 81] x^2 +  [16 b^4 - 120 b^3 + 324 b^2 - 378 b + 162] x^3 +
[- 16 b^4 + 96 b^3 - 216 b^2 + 216 b - 81] x^4
$$
We have the following signs, symbolically denoted by
$$
F(x)  = ["+"] + ["+"]  x + ["-"] x^2 +  ["+"] x^3 +
["-"] x^4
$$
for all $b \in [1 \quad \sqrt[4]{2} ]$, which makes three sign changes in the coefficients of $F(x)$.
For $x+1$, this gives
$$
F(x+1)  = [- 4 b^6 + 12 b^5 - 18 b^4 + 24 b^3 - 16 b^2 + 2] +[ - 8 b^5 + 24 b^4 - 54 b^3 + 126 b^2 - 144 b + 54] b x + [- 4 b^6 + 12 b^5 - 69 b^4 + 276 b^3 - 459 b^2 + 324 b - 81] x^2 +  [- 48 b^4 + 264 b^3 - 540 b^2 + 486 b - 162] x^3 +
[- 16 b^4 + 96 b^3 - 216 b^2 + 216 b - 81] x^4
$$
Again we have the following signs, symbolically denoted by
$$
F(x+1)  = ["+"] + ["-"]  x + ["-"] x^2 +  ["+"] x^3 +
["-"] x^4
$$
for all $b \in [1 \quad \sqrt[4]{2} ]$, which makes again three sign changes in the coefficients of $F(x+1)$.
This establishes that for all $b$, the polynomial $F$ has no roots, which completes the proof. $\qquad  \Box$
** Remark: For the two ranges of $b$, another proof of the inequality  $F\ge 0$ with the help of Buffalo Way has been given by River Li here **