Inequality

x=\sqrt{3}tan(\frac{A}{2}) \\ y=\sqrt{3}tan(\frac{B}{2}) \\ z=\sqrt{3}tan(\frac{C}{2}) \\ \sum{x\sqrt{x^2 + \frac{1}{x^2}}}\geq \sum{x}\sqrt{2}\geq \sqrt{3}\sqrt{2}\sum{tan\frac{A}{2}}\geq 3\sqrt{2}

x⁴ , y⁴ , z⁴ , are always +ve , independant of x , y , z .

So we can apply am-gm on 1 + x⁴ , etc.

We get 1 + x⁴2 ≥ x²

or √1 + x⁴ ≥ √2 x

similarly √1 + y⁴ ≥ √2 y

and √1 + z⁴ ≥ √2 z

adding all of them together , we get L.H.S ≥ √2 ( x + y + z) ---------------- 1

now ( x+ y + z )² = x² + y² + z² + ( xy + zx + yz ) ≥ 9 ,--------------A

as given xy + zx + yz = 3 and accordingly each of x² , y² , z² ≥ 1 ( if none of them are zero )

however , even if one or two of x , y , z are zero , then also we get either two or one of x , y , z ≥ 2

so from A , we get x + y + z ≥ 3 ------------------ 2

again from 1 and 2 , L.H.S ≥ √2 ( 3 ) ≥ 3√2

As started by Soumya, we arrive at \sqrt{1+x^4}\ge \sqrt{2}x, \sqrt{1+y^4}\ge \sqrt{2}y, \sqrt{1+z^4}\ge \sqrt{2}z

(x+y+z)^2=x^2+y^2+z^2+6

From Cauchy-Schwarz on (a_1,a_2,a_3)=(x,y,z) & (b_1,b_2,b_3)=(y,z,x), we arrive at (x^2+y^2+z^2)^2\ge 9 \Rightarrow (x+y+z)^2\ge 9

Now if x+y+z ≥ 3, it's as soumya has done....

But if (x+y+z)\le -3, then we write the entire thing - but this time taking the negative square roots like
\sqrt{1+x^4}\ge -\sqrt{2}x

Adding the 3 expressions we have our result.

I did using CS

2(1+x^4) \ge (1+x^2)^2 \Rightarrow \sqrt{2(1+x^4)} \ge 1+x^2

and we have \sum 1+x^2 = 3 + \sum x^2 \ge 3 + \sum xy = 6