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.
 A: It can be seen that $\sqrt{x^2+\frac{1}{x^2}}$ is convex. Therefore we can use Jensen inequality: 
$$
\frac{1}{3}\sqrt{a^2+1/a^2}+\frac{1}{3}\sqrt{b^2+1/b^2}+\frac{1}{3}\sqrt{c^2+1/c^2} \ge \sqrt{(\frac{a+b+c}{3})^2+(\frac{3}{a+b+c})^2}.
$$
On the other hand because $a+b+c<1$, we can see that $f(x)=\sqrt{\frac{x^2}{9}+\frac{9}{x^2}}$ for $x<1$ has its minimum at $x=1$ and hence:
$$
\frac{1}{3}\sqrt{a^2+1/a^2}+\frac{1}{3}\sqrt{b^2+1/b^2}+\frac{1}{3}\sqrt{c^2+1/c^2} \ge \sqrt{\frac{82}{9}}.
$$
which proves the inequality.
A: Assume first $a+b+c=1$.
Let $x=(a,1/a)$, $y=(b,1/b)$, $z=(1/c)$. Then the triangle inequality for the Euclidean norm tells us that 
$$
\|x+y+z\|\leq\|x\|+\|y\|+\|z\|.
$$
This looks like
$$
\sqrt{1+\left(\frac1a+\frac1b+\frac1c\right)^2}\leq\sqrt{a^2+\frac1{a^2}}+\sqrt{b^2+\frac1{b^2}}+\sqrt{c^2+\frac1{c^2}}.
$$
So to prove the inequality in the question we need to show that
$$
\frac1a+\frac1b+\frac1c\geq9.
$$
Using the Cauchy-Schwarz inequality:
$$9=\left\|( \sqrt{a}, \sqrt{b}, \sqrt{c})\cdot(\frac{1}{\sqrt{a}}, \frac{1}{\sqrt{b}}, \frac{1}{\sqrt{c}})\right\|^2\leq (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} . $$
Finally, when $a+b+c <1$, let $d=1-(a+b) $. Then $c <d <1$; as $\sqrt {x^2+1/x^2} $ is decreasing on $(0,1) $, we have $\sqrt {c^2+1/c^2}>\sqrt {d^2+1/d^2} $. Also, $a+b+d=1$ and so
$$
\sqrt{a^2+\frac1{a^2}}+\sqrt{b^2+\frac1{b^2}}+\sqrt{c^2+\frac1{c^2}}\geq
 \sqrt{a^2+\frac1{a^2}}+\sqrt{b^2+\frac1{b^2}}+\sqrt{d^2+\frac1{d^2}}\geq\sqrt {82}.
$$
A: In the first proof only condition $a+b+c=1$ is used. What if $a+b+c<1$? Also a simpler alternative way to show $\frac{1}{a}+ \frac{1}{b}+\frac{1}{c}\geq 9$ when $a+b+c\leq 1$ is the Cauchy-Schwarz inequality:
$9=\|( \sqrt{a}, \sqrt{b}, \sqrt{c})\cdot(\frac{1}{\sqrt{c}}, \frac{1}{\sqrt{b}}, \frac{1}{\sqrt{c}})\|^2\leq (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$.
A: 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.
