Problem. Find the smallest constant $K$ such that $$\sqrt{x}+\sqrt{y}+\sqrt{z}\le K\sqrt{xyz},$$ holds for all $x,y,z\ge 1: xy+yz+zx=2(x+y+z-1).$
I've tried to set $x=y=z=t$ and solve the equation: $3t^3-6t+2=0$ and I chose $t=\dfrac{3+\sqrt{3}}{3}.$
The inequality becomes $$K\ge \frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{xyz}}=\frac{3\sqrt{t}}{\sqrt{t^3}}=\frac{3}{t}=\dfrac{9-3\sqrt{3}}{2}.$$ Now, we will prove $$\sqrt{x}+\sqrt{y}+\sqrt{z}\le \dfrac{9-3\sqrt{3}}{2}\sqrt{xyz}.$$ I'm stuck here. The given condition $xy+yz+zx=2(x+y+z-1)$ is quite hard.
I hope to see some ideas. Thank you.
Let $x=1+a$, $y=1+b$ and $z=1+c$.
Thus, $a$, $b$ and $c$ are non-negatives and the condition gives: $$ab+ac+bc=1$$ and we need to find a maximal value of $$\sum_{cyc}\frac{1}{\sqrt{(a+1)(b+1)}}.$$ For $a=b=1$ and $c=0$ we obtain a value $\frac{2\sqrt2+1}{2}$.
We'll prove that it's a maximal value, for which it's enough to prove that: $$\sum_{cyc}\frac{1}{\sqrt{(a+1)(b+1)}}\leq\frac{2\sqrt2+1}{2}.$$ Indeed, let $c=0$.
Thus, $ab=1$ and by C-S and AM-GM we obtain: $$\sum_{cyc}\frac{1}{\sqrt{(a+1)(b+1)}}=\frac{1}{\sqrt{(a+1)(b+1)}}+\frac{1}{\sqrt{a+1}}+\frac{1}{\sqrt{b+1}}\leq$$ $$\leq\frac{1}{\sqrt{(a+1)(b+1)}}+\sqrt{(1+1)\left(\frac{1}{a+1}+\frac{1}{b+1}\right)}=$$ $$=\frac{1}{\sqrt{a+b+2}}+\sqrt{\frac{2(a+b+2)}{a+b+2}}\leq\frac{1}{\sqrt{2+2}}+\sqrt2=\frac{2\sqrt2+1}{2}.$$ Id est, it's enough to prove our inequality in the inside maximal point $(a,b,c)$.
Now, let $$f(a,b,c,\lambda)=\sum_{cyc}\sqrt{a+1}-\frac{2\sqrt2+1}{2}\sqrt{\prod_{cyc}(a+1)}+\lambda(ab+ac+bc-1).$$ Thus, $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=0$$ or $$\frac{1}{2\sqrt{a+1}}-\frac{2\sqrt2+1}{2}\frac{\sqrt{(b+1)(c+1)}}{2\sqrt{a+1}}+\lambda(b+c)=$$ $$=\frac{1}{2\sqrt{b+1}}-\frac{2\sqrt2+1}{2}\frac{\sqrt{(a+1)(c+1)}}{2\sqrt{b+1}}+\lambda(a+c)=$$ $$=\frac{1}{2\sqrt{c+1}}-\frac{2\sqrt2+1}{2}\frac{\sqrt{(b+1)(a+1)}}{2\sqrt{c+1}}+\lambda(b+a)=0.$$ Now, let $a\neq b$ and $a\neq c$.
Thus, $$\sqrt{a+1}+2\lambda(b+c)(a+1)=\sqrt{b+1}+2\lambda(a+c)(b+1)$$ or $$\frac{a-b}{\sqrt{a+1}+\sqrt{b+1}}+2\lambda(a-b)(c-1)=0$$ or $$\frac{1}{\sqrt{a+1}+\sqrt{b+1}}+2\lambda(c-1)=0.$$ By the similar way we obtain: $$\frac{1}{\sqrt{a+1}+\sqrt{c+1}}+2\lambda(b-1)=0,$$ which gives $$\left(\sqrt{a+1}+\sqrt{b+1}\right)(c-1)=\left(\sqrt{a+1}+\sqrt{c+1}\right)(b-1)$$ or $$b\sqrt{a+1}-c\sqrt{a+1}+b\sqrt{c+1}-c\sqrt{b+1}+\sqrt{b+1}-\sqrt{c+1}=0$$ or $$(b-c)\left(\sqrt{a+1}+\frac{bc+b+c}{b\sqrt{c+1}+c\sqrt{b+1}}+\frac{1}{\sqrt{b+1}+\sqrt{c+1}}\right)=0,$$ which gives $b=c$ and it's enough to prove our inequality for equality case of two variables.
Let $b=a$ and $c=\frac{1-a^2}{2a},$ where $0<a<1$.
Thus, we need to prove that: $$\frac{1}{a+1}+\frac{2}{\sqrt{(a+1)\left(\frac{1-a^2}{2a}+1\right)}}\leq\frac{2\sqrt2+1}{2}$$ and the rest is smooth.