Integral Inequality

[man111]

How can i prove [tex]\displaystyle \frac{3}{4}<\int^{1}_{0}\frac{1}{1+x^4}dx<1[/tex], please help me

Thanks

[nikola.topalov]

There is a similar inequality here: https://www.stumblingrobot.com/2015/12/ ... of-1-1-x4/

[Davids]

How can i prove [tex]\displaystyle \frac{3}{4}<\int^{1}_{0}\frac{1}{1+x^4}dx<1[/tex], please help me

Thanks

Well, compared to the results in the reply above, I found my approach rather simple, but here's what I think...

1. First, we'll consider the RHS:
$\int_0^1\frac{1}{1+x^4}dx < 1$

Now, since $1 = \int_0^1dx$, then we can express the inequality as follows:
$\int_0^1\frac{1}{1+x^4}dx < \int_0^1dx$
$\int_0^1\left(\frac{1}{1+x^4} - 1\right)dx < 0$
$\int_0^1\frac{-x^4}{1+x^4}dx < 0$

Which is obviously true, since our integrand is negative (not only for $x \in [0, 1]$, but also over its whole domain).

2. Now let's proceed with the LHS:
$\int_0^1\frac{1}{1+x^4}dx > \frac{3}{4}$

For that purpose, we can simply observe that for $x\in(0, 1)$:
$x^4 < x^2 \Longrightarrow \frac{1}{1+x^4} > \frac{1}{1 + x^2}$
$\Longrightarrow \int_0^1\frac{1}{1+x^4}dx > \int_0^1\frac{1}{1+x^2}dx = \arctan(x)\Bigg|_0^1 = \arctan(1) - \arctan(0) = \frac{\pi}{4} > \frac{3}{4}$

Which concludes the proof. But yet again, this seemed rather simpler than I expected, so... if something appers to be off, let me know