Maximum Value for Real x

Let $E = \frac{x}{\sqrt{1+x^4}}$. For $x \le 0$, $E \le 0$. To maximize, take $x > 0$. $E = \frac{1}{\sqrt{\frac{1}{x^2} + x^2}}$ By AM-GM inequality, $x^2 + \frac{1}{x^2} \ge 2$. Thus, $\sqrt{x^2 + \frac{1}{x^2}} \ge \sqrt{2}$. Therefore, $E \le \frac{1}{\sqrt{2}}$. The maximum value occurs when $x^2 = 1 \implies x = 1$, yielding $\frac{1}{\sqrt{2}}$.