Let's determine the values of the parameter $a$ for which the inequality

$$
\sqrt{1+x^{4}}>1-2 a x+x^{2}
$$

holds for all positive $x$.
Solution. We can divide the inequality by $x$, since $x>0$:

$$
\sqrt{\frac{1}{x^{2}}+x^{2}}>\frac{1}{x}+x-2 a
$$

Further transformations yield

$$
2 a>\frac{1}{x}+x-\sqrt{\frac{1}{x^{2}}+x^{2}}=\frac{1}{x}+x-\sqrt{\left(x+\frac{1}{x}\right)^{2}-2}
$$

Let's denote $x+\frac{1}{x}$ by $y$, then

$$
\begin{aligned}
2 a>y-\sqrt{y^{2}-2} & =\frac{\left(y-\sqrt{y^{2}-2}\right)\left(y+\sqrt{y^{2}-2}\right)}{y+\sqrt{y^{2}-2}}= \\
& =\frac{y^{2}-y^{2}+2}{y+\sqrt{y^{2}-2}}=\frac{2}{y+\sqrt{y^{2}-2}}
\end{aligned}
$$

which implies

$$
a>\frac{1}{y+\sqrt{y^{2}-2}}
$$

Considering that the minimum value of $y=x+\frac{1}{x}$ is $2$, $\sqrt{y^{2}-2} \geq \sqrt{2}$, and

$$
\frac{1}{y+\sqrt{y^{2}-2}} \leq \frac{1}{2+\sqrt{2}}=1-\frac{\sqrt{2}}{2}
$$

Thus, $1-\frac{\sqrt{2}}{2} \geq \frac{1}{y+\sqrt{y^{2}-2}}$. Therefore, the upper bound (the least upper bound) is $1-\frac{\sqrt{2}}{2}$, so the values of the parameter $a$ that satisfy the condition are those for which $a>1-\frac{\sqrt{2}}{2}$.