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!
 A: Yes it's Ok, you could also just multiply both sides by $2(1+x^4)$ having to prove that
$$
2x^2\le x^4+1
$$ that is
$$
0\le(x^2-1)^2.
$$
A: 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 $x\in\Bbb R$.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
\color{#f00}{x^{2} \over 1 + x^{4}} & =
{1 \over 1/x^{2} + x^{2}} =
{1 \over \pars{1/x - x}^{2} + 2} \color{#f00}{\leq \half}
\end{align}

The equality is satisfied whenever $\ds{1/x - x = 0\quad\imp\quad x = \pm 1}$.

