Prove $\sqrt{a + ab} + \sqrt{b} + \sqrt{c} \ge 3$ for $c = \min(a, b, c)$ and $ab + bc + ca = 2$

Question

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.

Answer 1

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 **

Answer 2

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.