<s>[INST] Does this constitute a valid proof that $\frac{x^2}{1+x^4} \leq \frac{1}{2}$? 
Prove that $$\frac{x^2}{1+x^4} \leq \frac{1}{2}.$$

First off, we observe that the expression on the LHS is positive for all $x \in \Bbb R,$ and equality is achieved iff $x \in \{-1, 1 \}$. That being said, we start by manipulating the expression as below;
\begin{align}
\frac{x^2}{1+x^4} - \frac{1}{2} &\leq 0\\\\
\impliedby\frac{2x^2 - (1+x^4)}{2(1+x^4)} &\leq 0\\\\
\impliedby -\frac{(x^4 - 2x^2 + 1)}{2(1+x^4)}&\leq 0\\\\
\impliedby \frac{x^4 - 2x^2 + 1}{2(1+x^4)} &\geq 0\\\\
\end{align} 
We then make the observation that $x^4 - 2x^2 + 1 = (x^2 - 1)^2 \geq 0$ for all $x \in \Bbb R$ and that $1 + x^4 \geq 1$ for all $x \in \Bbb R$, and so the final inequality holds for all $x \in \Bbb R. \\\hspace{92pt}\square$
Is it okay to prove it in this way? I have no solutions so it would be nice if I could have some validation!
 [/INST] Start from
$$x^4-2x^2+1=(x^2-1)^2\ge0$$
which is true for all $x\in\Bbb R$. Then rearrange:
$$2x^2\le1+x^4$$
$$\frac{x^2}{1+x^4}\le\frac12$$
where the last step is done because 2 and $1+x^4$ are always positive for $x\in\Bbb R$.
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