ถ้า $a,b,c \geq 0$ นะครับ
$-(ab+bc+ca) \leqslant 0 ...(1)$
From Cauchy-Schwarz Inequality
$a\sqrt{a-b}\sqrt{a-b}+b\sqrt{b-c}\sqrt{b-c}+c\sqrt{c-a}\sqrt{c-a}$
$\leqslant \sqrt{a^2(a-b)+b^2(b-c)+c^2(c-a)}\sqrt{a-b+b-c+c-a}=0$
$a(a-b)+b(b-c)+c(c-a) \leqslant 0 ...(2)$
$(1)+(2);a(a-b)+b(b-c)+c(c-a)-(ab+bc+ca) \leqslant 0$
$a^2+b^2+c^2-2(ab+bc+ca) \leqslant 0$
$3(a^2+b^2+c^2)-6(ab+bc+ca) \leqslant 0$
$12(a^2+b^2+c^2+ab+bc+ca)-9(a^2+b^2+c^2+2ab+2bc+2ca) \leqslant 0$
$12(a^2+b^2+c^2+ab+bc+ca) \leqslant 9(a^2+b^2+c^2+2ab+2bc+2ca)$
$3(a^2+b^2+c^2+ab+bc+ca) \leqslant \frac{9}{4}(a+b+c)^2 ...(3)$
From Cauchy-Schwarz Inequality
$2(\sqrt{a^2+bc}+\sqrt{b^2+ca}+\sqrt{c^2+ab}) \leq 2(\sqrt{3}\sqrt{a^2+b^2+c^2+ab+bc+ca})$
$=2(\sqrt{3(a^2+b^2+c^2+ab+bc+ca)})$
$\leq 2(\sqrt{\frac{9}{4}(a+b+c)^2}) (from(3))$
$=3(a+b+c)$
$\therefore 2(\sqrt{a^2+bc}+\sqrt{b^2+ca}+\sqrt{c^2+ab}) \leq 3(a+b+c)$
