The sum of three square roots bounded below by $\sqrt{82}$ 
Let $a,b,c >0$ and $a+b+c \le  1$. Prove: $$\sqrt{a^2+1/a^2}+\sqrt{b^2+1/b^2}+\sqrt{c^2+1/c^2} \ge \sqrt{82}$$

Progress: I tried to have 3 vectors $(a,1/a)$, $(b,1/b)$ and $(c,1/c)$, played around with the vector inequalities but no success.

since $f（x)=x^2+\dfrac{1}{x^2}$ is mono decreasing function when $x \in (0,1] $, we can prove $a+b+c=1$ case first, for any $a'+b'+c' <1$, let $p(a'+b'+c')=1 \implies p>1$, then $a=pa'>a',b=pb'>b',c=pc'>c' \implies \sum \sqrt{a'^2+\dfrac{1}{a'^2}}>\sum\sqrt{a^2+\dfrac{1}{a^2}}$
to prove the case $a+b+c=1$, it is same as Martin's post:
$\sum \sqrt{x_i^2+y_i^2} \ge \sqrt{(\sum x_i)^2+ (\sum y_i)^2}$,that is:
LHS$\ge \sqrt{(a+b+c)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}$
$\dfrac{3}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}} \le \dfrac{a+b+c}{3} \implies \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \ge 9$
QED.
