For real x, the maximum possible value of $\frac{x}{\sqrt{1+x^4}}$ is

The correct answer is $\frac{1}{\sqrt{2}}$.

We need to find the maximum value of the function:

$f\left(x\right)=\frac{x}{\sqrt{1+x^4}}$

Notice that if $x<0$, the function is negative, and if $x=0$, the function equals 0. So the maximum must occur for some $x>0$. We restrict our search to positive real values of $x$.

Step 1: Square the function to simplify the optimization.

Since $x>0$, maximizing $f\left(x\right)$ is equivalent to maximizing its square:

$g\left(x\right)={\left[f\right(x\left)\right]}^2=\frac{x^2}{1+x^4}$

Step 2: Apply the AM–GM Inequality.

The Arithmetic Mean – Geometric Mean (AM–GM) Inequality states that for any two positive real numbers $a$ and $b$:

$\frac{a+b}{2}\ge \sqrt{ab}$

which rearranges to:

$a+b\ge 2\sqrt{ab}$

Apply this to the denominator of $g\left(x\right)$, with $a=1$ and $b=x^4$:

$1+x^4\ge 2\sqrt{1\cdot x^4}=2x^2$

(valid since $x>0$, so $x^2>0$)

Step 3: Derive the upper bound for g(x).

Since $1+x^4\ge 2x^2$, taking reciprocals flips the inequality (both sides are positive):

$\frac{1}{1+x^4}\le \frac{1}{2x^2}$

Multiplying both sides by $x^2$ (which is positive):

$\frac{x^2}{1+x^4}\le \frac{x^2}{2x^2}=\frac{1}{2}$

So we have established:

$g\left(x\right)\le \frac{1}{2}$

Step 4: Find when equality holds (when the maximum is achieved).

In AM–GM, equality holds if and only if $a=b$. Here that means:

$1=x^4⟹x=1$ (taking the positive real root)

Let's verify by substituting $x=1$:

$f\left(1\right)=\frac{1}{\sqrt{1+1^4}}=\frac{1}{\sqrt{2}}$

This confirms that the maximum value is indeed attained at $x=1$.

Step 5: Conclusion.

Since $g\left(x\right)\le \frac{1}{2}$, we get:

$f\left(x\right)=\sqrt{g\left(x\right)}\le \sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$

Therefore, the maximum possible value of $\frac{x}{\sqrt{1+x^4}}$ for real $x$ is:

$\frac{1}{\sqrt{2}}$