Factoring!

factoring the expression using the Sophie Germain identity and using the sum property of logarithms we have

$\ln { (x^{ 2 }+\sqrt { 2 } x+1) }=\ln{(x^2+\sqrt{2} x+1)}+\ln{(x^2-\sqrt{2} x+1)}$ , equality since the problem is asking for the smallest such x

subtracting LHS from both sides we have

$0=\ln{(x^2-\sqrt{2}x+1)}$

$\Rightarrow x^2-\sqrt{2}x=0 \Rightarrow x=0 or x=\sqrt{2}$ dismissing 0 since it's not positive

so we get our answer

$\boxed{x=\sqrt{2}}$

(squaring that we would get 2 which is the actual answer to the problem)

$x^{2}+\sqrt{2}x+1<x^{4}+1 => x^{4}-x^{2}-\sqrt{2}x>0 => x(x^{3}-x-\sqrt{2})>0$
Since x is positive, $x^{3}-x-\sqrt{2}>0$ . It is clear that $\sqrt{2}$ is a root of the cubic by hit and trial and thus minimum value = $\sqrt{2}$

We know that we can factor $x^4+4$ with integer factors.
$\therefore ~ Completing~ the~ sqare~x^4+1=(x^4 +2x^2+1) - 2x^2=(x^2+\sqrt2x+1)*(x^2 - \sqrt2x+1).\\ \therefore~0\leq Ln(x^4+1) - Ln (x^2+\sqrt2x+1)=Ln (x^2 - \sqrt2x+1), but~ Ln~ we ~have~ \nless~ 0.\\ \therefore~ Ln (x^2+\sqrt2x+1)=0 ~ \implies ~x^2+\sqrt2x =0, ~ x=0 ~ give ~ minimum, ~ x=\sqrt2 ~ maximum.\\ ~ ~ Ans=2 .$