Need to prove $\frac{3}{5}(2^{\frac{1}{3}}-1)\le\int_0^1\frac{x^4}{(1+x^6)^{\frac{2}{3}}}dx\le1$

Question

I need to show that $$\frac{3}{5}(2^{\frac{1}{3}}-1)\le\int_0^1\frac{x^4}{(1+x^6)^{\frac{2}{3}}}dx\le1$$ I just know that if in $[a,b]$, $f(x)\le g(x)\le h(x)$, then $$\int_a^bf(x)dx\le\int_a^bg(x)dx\le\int_a^bh(x)dx$$ How do I get a function like $f$ and $h$ in the range $[0,1]$, for the function $g(x)=\frac{x^4}{(1+x^6)^{\frac{2}{3}}}$? I mean, is there an approach without solving backwards?

Answer 1

Consider left part first. You need to get integral there, something like $f(1)-f(0)$ $$ \frac 35 \left ( 2^{\frac 13}-1\right ) = \frac 35 \left ( (1+\fbox 1)^{\frac 13}-(1+\fbox 0)^{\frac 13}\right ) =\frac 35 \left . \left ( 1+x^n\right )^{\frac 13} \right |_0^1 $$ Now, you can imagine that values inside of the boxes are values of almost any power of $x$, since $$1^n = 1 \\ 0^n = 0$$ So $f(x) = \frac 35 \left (1+x^n \right )^{\frac 13}$, and therefore $f'(x) = \frac n5 \left ( 1+x^n\right )^{-\frac 23}x^{n-1}$

To choose which power should be there, just pay attention to the integral given. It has $x^4$ in enumerator, and $(1+x^6)$ in denominator, so power could be either 5 (to match enumerator after differentiating) or 6 (to match denominator). $$ n = 5 \\ f(x) = \frac 35 \left (1+x^5 \right )^{\frac 13} \\ f'(x) = \frac {x^4}{\left ( 1+x^5\right)^{\frac 23}} $$ Now, all you have to prove is $$ \frac{x^4}{\left (1+x^5 \right)^{\frac 23}} \le \frac {x^4}{\left (1+x^6 \right)^{\frac 23}},\quad \text{for}\ \forall x \in [0,1] $$ which is equivalent to $$ 1 + x^6 \le 1 + x^5 $$ or $$ x \le 1 $$ which is true.

If you choose to match denominator $$ n = 6 \\ f(x) = \frac 35 \left ( 1+x^6\right )^{\frac 13} \\ f'(x) = \frac 65 \left ( 1+x^6\right)^{-\frac 23} x^5 $$ You have to prove that $$ \frac {6 x^5}{5\left ( 1+x^6\right)^{\frac 23}} \le \frac {x^4}{\left (1+x^6 \right)^{\frac 23}},\quad \text{for}\ \forall x \in [0,1] $$ which is equivalent to $$ 6x \le 5 $$ which is not true for all $x \in [0,1]$, so it's better to stick with $n = 5$.

Right part is easier, I can put some notes if it's necessary.

Answer 2

Try $\dfrac{x^4}{(1+x^5)^{2/3}}$