Find the minimum value of $P=ab+bc+3ca+\dfrac{3}{a+b+c}$ Let $a,b,c$ be non-negative real numbers such that $a^2+b^2+c^2=3$. Find the minimum value of $$P=ab+bc+3ca+\dfrac{3}{a+b+c}.$$
This is an asymmetric inequality. It is hard for me to find when the equation holds. I guess when it occurs if $a=c=0$ and $b=\sqrt{3}$. Then $\min P=\sqrt{3}$. But I have no idea to solve it. Please help me a hint. Thank you.

As $a,c \ge 0$ then
$$P \ge S := ab+bc+ca + \frac{3}{a+b+c}$$
We minimize $S$
As $a^2+b^2 +c^2 = 3$ then
$$S =\frac{1}{2} (a+b+c)^2+\frac{3}{a+b+c} -\frac{3}{2}$$
Denote $x = a+b+c$. Because $a^2+b^2 +c^2 = 3$ then $\sqrt{3} \le x \le 3$
$$S = f(x) := \frac{1}{2}x^2+\frac{3}{x}-\frac{3}{2}$$
As $f'(x) = x-\frac{3}{x^2}>0$ in the interval $[\sqrt{3},3]$ then $f(x)$ is increasing. Hence, $S_1 = f(x)$ is minimized as $a+b+c = x_0 = \sqrt{3}$
Conclusion: we can conclude that
$$P \ge  S \ge \sqrt{3} $$
The equality occurs when $(a,b,c) = (0,\sqrt{3},0),(0,0,\sqrt{3})$ or $(\sqrt{3},0,0)$
