Using the comparison test to evaluate $\int_1^\infty\frac{1}{1+x^2+16x^4}dx$?

Question

So using the comparison test to evaluate $\int_1^\infty\frac{1}{1+x^2+16x^4}dx$, and we're given $\int_1^\infty\frac{1}{4x^2}dx$. So I have been trying to set up an inequality to use, but I can't seem to turn the given integral into the original integral. This is what I've done thus far. $$\forall x(x\ge1)\rightarrow x\ge1 \rightarrow \hspace{10pt} 3x^2\ge x \ge 1$$ $$3x^2 \ge 1 \rightarrow \hspace{10pt} 3x^2+x^2 \ge 1+x^2$$ $$4x^2\ge1+x^2 \rightarrow\hspace{10pt}(4x^2)^2\ge(1+x^2)^2$$ $$16x^4\ge1+2x^2+x^4$$ And that's pretty much as far as I've gotten. I know $\frac{1}{1+x^2+16x^4} \le \frac{1}{4x^2} $, and that $\frac{1}{4x^2}$ converges, so the original integral does too. I just don't know how to correctly proof that inequality. Thanks for all the help in advance.

Answer 1

Just see this if $x\geq 1$

$$ x^2 \geq x \implies x^4 \geq x^2 \implies 16 x^4 \geq 4x^2 \implies 1+x^2+16 x^4 \geq 4x^2 $$

since you adding positive terms. So can you see it now?

Answer 2

By AM-GM $1+16x^4\geq 8x^2$, hence: $$\int_{1}^{+\infty}\frac{dx}{1+4x^2+16x^4}\leq\int_{1}^{+\infty}\frac{dx}{12x^2}.$$

Answer 3

Another (probably) easier solution $$ 1+x^2+16x^4 \geq 1+x^2 \implies \frac{1}{1+x^2+16x^4} \leq \frac{1}{1+x^2} \implies \\ \int_{1}^{+\infty}\frac{dx}{1+x^2+16x^4} \leq \int_{1}^{+\infty}\frac{dx}{1+x^2} = \frac{\pi}{4} $$