Old and easy!

Just AM-GM

Trivial by Minkowski. $\sum_{cyc} \sqrt{x^2+\frac{1}{y^2}} \ge \sqrt{(x+y+z)^2+(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2}$ Now, ${(x+y+z)^2+(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2} \ge \sqrt{18}=3\sqrt{2}$ (By AM-GM) (Done)

$\sqrt {x^2+\frac {1}{y^2}}+ \sqrt {y^2+\frac {1}{z^2}}+ \sqrt {z^2+\frac {1}{x^2}}\geq 3 \{\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{z^2}\right)\left(z^2+\frac{1}{x^2}\right)\}^\frac{1}{6}$ $\geq 3 \{2^3\}^{\frac{1}{6}}=3\sqrt{2}$ by just AM-GM

$$\sum \sqrt {x^2+\frac {1}{y^2}}\geq\sum \sqrt{\left |\frac{x}{y} \right |}\sqrt{2}\geq 3\sqrt{2}$$

WLOG $x,y,z>0$ $\displaystyle\sum\limits_{cyc}\sqrt{x^2+\dfrac{1}{y^2}}\ge \sqrt 2\sum\limits_{cyc}\sqrt{\dfrac{x}{y}}$ by AM-GM. Now make the substitution $a=\dfrac xy,b=\dfrac yz,c=\dfrac zx$. Then we wish to show $\displaystyle\sum\limits_{cyc} \sqrt a\ge 3$ for all $a,b,c>0$ satisfying $abc=1$ which is obvious by AM-GM.

easy but it is good problem

Dilshodbek wrote: easy but it is good problem

SohamSchwarz119 wrote: Trivial by Minkowski. $\sum_{cyc} \sqrt{x^2+\frac{1}{y^2}} \ge \sqrt{(x+y+z)^2+(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2}$ Now, ${(x+y+z)^2+(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2} \ge \sqrt{18}=3\sqrt{2}$ (By AM-GM) (Done) good solution

SohamSchwarz119 wrote: Trivial by Minkowski. $\sum_{cyc} \sqrt{x^2+\frac{1}{y^2}} \ge \sqrt{(x+y+z)^2+(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2}$ Now, ${(x+y+z)^2+(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2} \ge \sqrt{18}=3\sqrt{2}$ (By AM-GM) (Done) but x y z R

Dilshodbek wrote: SohamSchwarz119 wrote: Trivial by Minkowski. $\sum_{cyc} \sqrt{x^2+\frac{1}{y^2}} \ge \sqrt{(x+y+z)^2+(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2}$ Now, ${(x+y+z)^2+(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2} \ge \sqrt{18}=3\sqrt{2}$ (By AM-GM) (Done) but x y z R don't double post using editor

SOLUTION Simply use AM-GM twice. $$\sqrt {x^2+\frac {1}{y^2}}+ \sqrt {y^2+\frac {1}{z^2}}+ \sqrt {z^2+\frac {1}{x^2}} \geq 3\sqrt[6]{x^2y^2z^2+x^2+y^2+z^2+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}} \geq 3\sqrt[6]{8} = 3\sqrt{2}$$ PROBLEM SOURCE Azerbaijan Junior Mathematical Olympiad 2016

Jensen's works too. $$\frac{1}{3}\left(\sqrt {x^2+\frac {1}{y^2}}+ \sqrt {y^2+\frac {1}{z^2}}+ \sqrt {z^2+\frac {1}{x^2}}\right) \geq \sqrt{\frac{x^2+\frac {1}{x^2}+y^2+\frac {1}{y^2}+z^2+\frac {1}{z^2}}{3}} \geq \sqrt2$$$$\implies \sqrt {x^2+\frac {1}{y^2}}+ \sqrt {y^2+\frac {1}{z^2}}+ \sqrt {z^2+\frac {1}{x^2}}\geq 3\sqrt2$$