Finding $\small{\min\limits_{ab+bc+ca=1}\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}- \sqrt{2-abc}.}$

Question

Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=1.$ Find the minimal value of expression $$P=\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}- \sqrt{2-abc}.$$ By $a=b=1;c=0$ I get $P=2\sqrt{3}$ so we need to prove that$$\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}\ge \sqrt{2-abc}+2\sqrt{3}. $$ I thought of Lagrange multiplier method.

We can split inequality into two cases.

$\bullet$ If $a=0;bc=1,$ it's $\sqrt{b+2}+\sqrt{c+2}\ge 2\sqrt{3},$which is easy be AM-GM.

$\bullet$ $abc\neq0,$ we consider$$f(a,b,c,\lambda)=\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}-\sqrt{2-abc}-2\sqrt{3}+\lambda(ab+ac+bc-1).$$ and $(a,b,c)$ be an inside minimum point of $f$.

Thus, $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=0,$$ which gives \begin{align*} \frac{1}{2\sqrt{a+2}}+\frac{bc}{2\sqrt{2-abc}}+2\lambda(b+c)&=\frac{1}{2\sqrt{b+2}}+\frac{ac}{2\sqrt{2-abc}}+2\lambda(a+c)\\&=\frac{1}{2\sqrt{c+2}}+\frac{ba}{2\sqrt{2-abc}}+2\lambda(b+a). \end{align*} Now, let in this point $a\neq b$ and $a\neq c$ and we need to prove $b=c.$

From here, I was confusing in case $abc=2$ and how to continue my idea.

Hope you help me. Also, all idea and comment are welcome.

Update 1. Michael Rozenberg has almost solved it by LM. Is there another approach?

Answer 1

By your work $$\tfrac{1}{2\sqrt{a+2}}+\tfrac{bc}{2\sqrt{2-abc}}+\lambda(b+c)=\tfrac{1}{2\sqrt{b+2}}+\tfrac{ac}{2\sqrt{2-abc}}+\lambda(a+c)=\tfrac{1}{2\sqrt{c+2}}+\tfrac{ba}{2\sqrt{2-abc}}+\lambda(b+a)=0$$ or $$\tfrac{a}{2\sqrt{a+2}}+\tfrac{abc}{2\sqrt{2-abc}}+\lambda(ab+ac)=\tfrac{b}{2\sqrt{b+2}}+\tfrac{abc}{2\sqrt{2-abc}}+\lambda(ab+cb)=\tfrac{c}{2\sqrt{c+2}}+\tfrac{abc}{2\sqrt{2-abc}}+\lambda(ac+bc)=0$$ or $$\tfrac{a}{2\sqrt{a+2}}+\lambda(ab+ac)=\tfrac{b}{2\sqrt{b+2}}+\lambda(ab+cb)=\tfrac{c}{2\sqrt{c+2}}+\lambda(ac+bc)=-\tfrac{abc}{2\sqrt{2-abc}}.$$ Now, let in this point $a\neq b$ and $a\neq c$.

Thus, we have $$\tfrac{a}{2\sqrt{a+2}}+\lambda(ab+ac)=\tfrac{b}{2\sqrt{b+2}}+\lambda(ab+cb)$$ or $$\frac{a}{\sqrt{a+2}}-\frac{b}{\sqrt{b+2}}+2\lambda c(a-b)=0$$ or $$(a-b)\left(\frac{ab+2(a+b)}{\sqrt{(a+2)(b+2)}\left(a\sqrt{b+2}+b\sqrt{a+2}\right)}+2\lambda c\right)=0$$ or $$\frac{ab+2(a+b)}{\sqrt{(a+2)(b+2)}\left(a\sqrt{b+2}+b\sqrt{a+2}\right)}+2\lambda c=0$$ by the similar way we obtain $$\frac{ac+2(a+c)}{\sqrt{(a+2)(c+2)}\left(a\sqrt{c+2}+c\sqrt{a+2}\right)}+2\lambda b=0,$$ which gives $$\frac{b(ab+2(a+b))}{\sqrt{(a+2)(b+2)}\left(a\sqrt{b+2}+b\sqrt{a+2}\right)}=\frac{c(ac+2(a+c))}{\sqrt{(a+2)(c+2)}\left(a\sqrt{c+2}+c\sqrt{a+2}\right)}$$ or $$b(ab+2(a+b))\left(a(c+2)+c\sqrt{(a+2)(c+2)}\right)=$$ $$=c(ac+2(a+c))\left(a(b+2)+b\sqrt{(a+2)(b+2)}\right)$$ or $$a\left(b(c+2)(ab+2(a+b))-c(b+2)(ac+2(a+c))\right)+$$ $$+bc\sqrt{a+2}\left((ab+2(a+b))\sqrt{c+2}-(ac+2(a+c))\sqrt{b+2}\right)=0$$ or $$(b-c)\left(a(abc+2(ab+ac+bc)+4(a+b+c))+\tfrac{bc\sqrt{a+2}\left(a^2(bc+2b+2c+4)+4abc+4(2ab+2ac+bc)+8(2a+b+c)\right)}{(ab+2(a+b))\sqrt{c+2}+(ac+2(a+c))\sqrt{b+2}}\right)=0,$$ which gives $b=c$ and it's enough to prove our inequality for equality case of two variables.

Can you end it now?

Answer 2

Here is another approach. We assume $a,b,c$ are nonnegative. Let us denote the constraint set as $$C=\{(a,b,c): ab\leq 1, a\geq 0,b\geq 0, c=\frac{1-ab}{a+b} \}.$$ Note that $-\sqrt{2-abc}\geq -\sqrt{2} $. Thus $$\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}-\sqrt{2-abc} \geq$$ $$\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}-\sqrt{2}$$

Hence,
$$\min\limits_{(a,b,c)\in C} \sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}-\sqrt{2-abc} \geq$$ $$\min\limits_{(a,b,c)\in C} \sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}-\sqrt{2}$$ Let us define $f(a,b,c)=\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}-\sqrt{2}$ and $f(a,b,c)= f_1(a,b)+f_2(c)$ where $f_1(a,b)=\sqrt{a+2}+\sqrt{b+2}$ and $f_2(c)=\sqrt{c+2}-\sqrt{2}$.

Note that $$\min\limits_{(a,b,c)\in C}f(a,b,c)=\min\limits_{(a,b,c)\in C}f_1(a,b)+f_2(c)\geq$$ $$\min\limits_{(a,b,c)\in C}f_1(a,b)+\min\limits_{(a,b,c)\in C}f_2(c) \geq $$ $$\min\limits_{ab\leq 1 \& a,b,c\geq 0}f_1(a,b) + \min\limits_{ab\leq 1 \& a,b,c\geq 0}f_2(c)$$.

It is clear that $\min\limits_{ab\leq 1 \& a,b,c\geq 0}f_2(c)=0$ for $c=0$.

For $\min\limits_{ab\leq 1 \& a,b,c\geq 0}f_1(a,b)$, we have to find the minimum of $\sqrt{a+2}+\sqrt{b+2}$ for $a\geq 0, b>0, ab\leq 1$. Here, we can use Langrange multipliers and get the minimum at $a=b=1$. As a result we get $\min\limits_{ab\leq 1 \& a,b,c\geq 0}f_1(a,b)=2\sqrt{3}$.

This shows $\min\limits_{(a,b,c)\in C}f(a,b,c)\geq 2\sqrt{3}$. Since $(1,1,0)\in C$ and $f(1,1,0)=2\sqrt{3}$, we deduce that $\min\limits_{(a,b,c)\in C}f(a,b,c)= 2\sqrt{3}$.
Thus,
$$\min\limits_{(a,b,c)\in C} \sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}-\sqrt{2-abc} \geq \min\limits_{(a,b,c)\in C}f(a,b,c)= 2\sqrt{3} $$

So $\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}-\sqrt{2-abc}$ attains its minimum at $(1,1,0)$, which $2\sqrt{3}$ because $(1,1,0) \in C$.

Answer 3

Proof.

We need to prove that $$\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}\ge \sqrt{2-abc}+2\sqrt{3}.\tag{1}$$

WLOG, assume that $a \ge b \ge c$. We have $$\frac13 \le ab \le 1, \quad c \le \frac{1}{\sqrt{3}}. \tag{2}$$

By AM-GM, we have \begin{align*} \sqrt{a + 2} + \sqrt{b + 2} &= \sqrt{a + 2 + b + 2 + 2\sqrt{(a + 2)(b + 2)}}\\[6pt] &\ge \sqrt{a + 2 + b + 2 + 2\cdot \frac{2(a + 2)(b + 2)}{a + 2 + b + 2}}. \tag{3} \end{align*}

Using (2), we have $$\sqrt{c + 2} - \sqrt{2} = \frac{c}{\sqrt{c + 2} + \sqrt{2}} \ge \frac{c}{\sqrt{1/\sqrt{3} + 2} + \sqrt{2}} \ge \frac{50}{151}c. \tag{4}$$

We have $$\sqrt{2} - \sqrt{2 - abc} = \frac{abc}{\sqrt{2} + \sqrt{2 - abc}} \ge \frac{abc}{\sqrt{2} + \sqrt{2}} \ge \frac{6}{17}abc. \tag{5}$$

From (1) and (3)-(5), it suffices to prove that $$\sqrt{a + 2 + b + 2 + 2\cdot \frac{2(a + 2)(b + 2)}{a + 2 + b + 2}} + \frac{50}{151}c + \frac{6}{17}abc \ge 2\sqrt{3}. \tag{6}$$

Let $p = a + b, q = ab$. We have $$1/3 \le q \le 1, \quad p^2 \ge 4q.$$

Using $c = \frac{1 - ab}{a + b} = \frac{1 - q}{p}$, (6) is written as $$f(q) := \sqrt{p + 4 + \frac{4(q + 2p + 4)}{p + 4}} + \frac{50}{151}\cdot \frac{1 - q}{p} + \frac{6}{17}\cdot q \cdot \frac{1 - q}{p} - 2\sqrt{3} \ge 0. \tag{7}$$

It suffices to prove that $f(q) \ge 0$ for $1/3 \le q \le \min(1, p^2/4)$.

Clearly, $f(q)$ is concave.

If $p\le 2$, it is not difficult to prove that $f(1/3) \ge 0$ and $f(p^2/4) \ge 0$. Thus, $f(q) \ge 0$ for all $q \in [1/3, p^2/4]$.

If $p > 2$, it is not difficult to prove that $f(1/3) \ge 0$ and $f(1) \ge 0$. Thus, $f(q)\ge 0$ for all $q \in [1/3, 1]$.

We are done.

Answer 4

Another way.

For $k=\sqrt{\frac{8}{3}}-1$ by Holder we obtain: $$\left(\sum_{cyc}\sqrt{a+2}\right)^2\sum_{cyc}(a+2)^2(ka+b+c)^3\geq\left(\sum_{cyc}(a+2)(ka+b+c)\right)^3$$ and it's enough to prove that: $$\left(\sum_{cyc}(a+2)(ka+b+c)\right)^3\geq\left(\sqrt{2-abc}+2\sqrt3\right)^2\sum_{cyc}(a+2)^2(ka+b+c)^3,$$ which save the case of the equality occurring.

Now, we can try to use $uvw$.

I hope it will help.