Áp dụng bđt Bunhiacốpski, ta có:$(\frac{1}{\sqrt{a^2+b^2+c^2}}.\sqrt{a^2+b^2+c^2}+\frac{1}{\sqrt{ab}}.3\sqrt{ab}+\frac{1}{\sqrt{bc}}.3\sqrt{bc}+\frac{1}{\sqrt{ca}}.\sqrt{ca})^2$
$\leq(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})(a^2+b^2+c^2+9ab+9bc+9ca)$
$\Rightarrow(1+3+3+3)^2 \leq (\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})[(a+b+c)^2+7(ab+bc+ca)]$
$\Rightarrow100\leq (\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})[(a+b+c)^2+\frac{7(a+b+c)^2}{3}$]
$\Rightarrow100 \leq(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}).\frac{10}{3}$ (do $a+b+c=1$)
$\color{red}{\Rightarrow\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\geq 30}$
Đẳng thức xảy ra khi và chỉ khi $\color{red}{a=b=c=\frac{1}{3}.}$