<s>[INST] 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.
 [/INST] 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
$$
\sqrt{a^2+\frac1{a^2}}+\sqrt{b^2+\frac1{b^2}}+\sqrt{c^2+\frac1{c^2}}\geq
 \sqrt{a^2+\frac1{a^2}}+\sqrt{b^2+\frac1{b^2}}+\sqrt{d^2+\frac1{d^2}}\geq\sqrt {82}.
$$
</s>