<s>[INST] Max of $3$-Variable Function 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 $\max_{a,b,c}f(a,b,c)=1-\frac{\sqrt{3}}{3}$
 [/INST] 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. 
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