TITLE: Find minimum of $\sqrt{a}+\sqrt{b}+\sqrt{c}$, given that $a,b,c \ge 0$ and $ab+bc+ca+abc=4$
QUESTION [8 upvotes]: Find the minimum of $\sqrt{a}+\sqrt{b}+\sqrt{c}$, given that $a,b,c \ge 0$ and $ab+bc+ca+abc=4$.

I can prove that $\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1,$ because:
$$ab+bc+ca+abc=4 \\ \implies (a+2)(b+2)(c+2)=(a+2)(b+2)+(b+2)(c+2)+(c+2)(a+2),$$
and $3 \le a+b+c$ (using $\frac{9}{a+b+c+6} \le \frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}$). But I didn't know how to deal with the $\sqrt{a}+\sqrt{b}+\sqrt{c}$.
Can anyone help me? Thank you so much.

REPLY [2 votes]: Not beautiful solution.
WLOG $a\leq b\leq c$. Suppose $a>1$, then $b>1$, $c>1$, $ab+bc+ca+abc > 4$. Therefore $a\leq 1$.
Let $\sqrt{b}+\sqrt{c}=d$, $\sqrt{bc}=e$. Then $bc=e^2$, $d^2=b+c+2e$, $b+c=d^2-2e$. $(\sqrt{b}-\sqrt{c})^2=b+c-2e=d^2-4e\geq 0$, therefore $d^2\geq 4e$.
$$ab+bc+ca+abc=4\Rightarrow a(b+c+bc)+bc=4 \Rightarrow a(d^2-2e+e^2)+e^2=4$$
$$d^2\geq 4e \Rightarrow a(4e-2e+e^2)+e^2=\leq 4 \Rightarrow (a+1)e^2+2ae-4\leq 0$$
$$(a+1)e^2+2ae-4=(a+1)e^2+2(a+1)e-2e-4=((a+1)e-2)(e+2)$$
$$(a+1)e^2+2ae-4\leq 0\Rightarrow (a+1)e-2\leq 0 \Rightarrow e\leq\frac2{a+1}$$
$$a(d^2-2e+e^2)+e^2=4\Rightarrow ad^2=4+2ae-(a+1)e^2=4+\frac{a^2}{a+1}-(a+1)\left(e-\frac{a}{a+1}\right)^2$$
$$0\leq e \leq \frac2{a+1}, a\leq 1\Rightarrow \left|e-\frac{a}{a+1}\right|\leq \frac{2-a}{a+1}\Rightarrow ad^2\geq 4+\frac{a^2}{a+1}-(a+1)\left(\frac{2-a}{a+1}\right)^2\Rightarrow$$
$$ad^2\geq \frac{8a}{a+1} \Rightarrow d^2\geq \frac8{a+1}\Rightarrow d\geq\frac{2\sqrt{2}}{\sqrt{a+1}}$$
$$\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{a}+d\geq \sqrt{a}+\frac{2\sqrt{2}}{\sqrt{a+1}}$$
Let $f=\sqrt{a}+\frac{2\sqrt{2}}{\sqrt{a+1}}$. Then $\sqrt{a}+\sqrt{b}+\sqrt{c}\geq f$.
$$f^2=a+\frac{8}{a+1}+\frac{4\sqrt{2a}}{\sqrt{a+1}}=a+8-\frac{8a}{a+1}+\frac{4\sqrt{2a}}{\sqrt{a+1}}-1+1=8+a+1-\left(2\sqrt{\frac{2a}{a+1}}-1\right)^2$$
$$a\leq 1 \Rightarrow a+a\leq a+1\Rightarrow 2\sqrt{\frac{2a}{a+1}}\leq 2\Rightarrow \left|2\sqrt{\frac{2a}{a+1}}-1\right|\leq 1\Rightarrow$$
$$f^2\geq 8+a+1-1=8+a\Rightarrow \sqrt{a}+\sqrt{b}+\sqrt{c}\geq f\geq \sqrt{8+a}\geq\sqrt{8}=2\sqrt{2}$$
Minimum is obtained at $a=0$ and $b=c$, because it requires $8+a=8$ and $\left|e-\frac{a}{a+1}\right|=\frac{2-a}{a+1}\Rightarrow e=\frac{2}{a+1}\Rightarrow d^2=4e\Rightarrow (\sqrt{b}-\sqrt{c})^2=0$.