<s>[INST] Doubt with an inequality involving integrals I have to prove this inequality about integrals. I did it but I'm not sure if my arguments were correct. Take a look:

Prove:
  $$0\leq\int_{-1}^1\frac{1-x^2}{x^4+1}dx\leq\int_{-1}^1\frac{2+x^4}{x^4+1}dx$$

My try is this:
Let $f(x)=\frac{1-x^2}{x^4+1}$ with $f:[-1,1]\to\mathbb{R}$ and $g(x)=\frac{2+x^4}{x^4+1}$ with $g:[-1,1]\to\mathbb{R}$. So we have to prove: $f(x)\leq g(x)$ for all $x\in[-1,1]$.
Clearly we only need to prove that $1-x^2\leq2+x^4$ because the denominator is always a positive number in $[-1,1]$. So we get:
$$0\leq x^2(x^2+1)+1=(x^2+1)(x^2+\frac{1}{x^2+1})$$
Therefore we need:
$$x^2+1\geq0 \land x^2+\frac{1}{x^2+1}\geq0$$
$$\lor x^2+1\leq0 \land x^2+\frac{1}{x^2+1}\leq0$$ 
The first pair of inequalities are always true for any $x\in\mathbb{R}$, because they involve positive numbers. And the second pair is the empty set by the same reason. 
Therefore $f(x)\leq g(x)$ for any $x\in\mathbb{R}$ in particular for $x\in[-1,1]$ so we have: $\int_{-1}^1\frac{1-x^2}{x^4+1}dx\leq\int_{-1}^1\frac{2+x^4}{x^4+1}dx$
The part of the zero is easy because we only need to prove: $0\leq1-x^2$, that is $x\leq|1|$, $-1\leq x\leq1$.
Is the first part correct or it is "forced"? Thanks!
 [/INST] Alt. hint: $\displaystyle\;\;\int_{-1}^1\frac{-x^2}{x^4+1}dx \lt 0 \lt \int_{-1}^1\frac{x^4}{x^4+1}dx\,$,  and $\displaystyle\;\int_{-1}^1\frac{1}{x^4+1}dx \lt \int_{-1}^1\frac{2}{x^4+1}dx\,$.
</s>