I'm trying the find the maximum of the function
$$f(a,b,c)=\frac{a+b+c-\sqrt{a^2+b^2+c^2}}{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}$$
for all nonnegative real numbers $a, b, c$ with $ab + bc + ca > 0$.
I tried in vain to prove that $\max_{a,b,c}f(a,b,c)=1-\frac{\sqrt{3}}{3}$
On
WLOG,let $c=$Min{$a,b,c$},
$f(a,b,c) \le \dfrac{a+b-\sqrt{a^2+b^2}}{\sqrt{ab}}=\dfrac{2ab}{\sqrt{ab}(a+b+\sqrt{a^2+b^2})} \le \dfrac{2ab}{\sqrt{ab}(2\sqrt{ab}+\sqrt{2ab})}=2-\sqrt{2}$
$\dfrac{a+b+c-\sqrt{a^2+b^2+c^2}}{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}} \le \dfrac{a+b-\sqrt{a^2+b^2}}{\sqrt{ab}} \iff \sqrt{ab}(a+b+c-\sqrt{a^2+b^2+c^2})\le (\sqrt{ab}+\sqrt{bc}+\sqrt{ca})(a+b-\sqrt{a^2+b^2}) \iff \sqrt{abc}((a^2+b^2)\sqrt{a}+(a^2+b^2)\sqrt{b}+c(a\sqrt{a}+b\sqrt{b}+a\sqrt{b}+b\sqrt{a})-2ab\sqrt{c}-\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})) \ge 0$
"=" will be hold only when $c=0$
so the max of $f(a,b,c)$ is $2-\sqrt{2}$ when $a=b,c=0$ or cycle.
the $1-\dfrac{\sqrt{3}}{3}$ is a paddle point.
You can assume $a +b +c = 3$ because scaling a solution by a positive number doesn't change the objective function. Then you are left with
$$\frac{3 - \sqrt{a^2 + b^2 + c^2}}{\sqrt{ab} + \sqrt{ac} + \sqrt{bc}}$$
subject to $a + b + c = 3$. Show that $a^2 + b^2 + c^2$ is minimized when $a=b=c$ assuming $a+b+c$ is a constant and similarly show that $\sqrt{ab} + \sqrt{ac} + \sqrt{bc}$ is minimized when $a=b=c$ assuming $a+b+c$ is a constant. Then you will get $a=b=c=1$ as an optimal solution, which gives the optimal value you quoted.