<s>[INST] Is there integral or series for $\sqrt{10}-\frac{4^4}{3^4}$ (to prove the inequality)? Both of these numbers are bad approximations for $\pi$, but they turn out to be much closer together:
$$\sqrt{10}-\frac{4^4}{3^4}=0.00178$$
Since there is a lot of questions here about integrals and series which prove such close inequalities, I wanted to know if something exists for this inequality as well:
$$\sqrt{10}>\frac{4^4}{3^4}$$
If we use the continued fraction for $\sqrt{10}$:
$$\sqrt{10}=3+\cfrac{1}{6+\cfrac{1}{6+\cdots}}$$
One of the approximants will be:
$$\sqrt{10} \approx \frac{117}{37}=\frac{234}{74}$$
$$\frac{4^4}{3^4}=\frac{256}{81}$$
This is just a curiosity for me, there is no other context for the question.
 [/INST] Well, this is the same as $\sqrt{2}\sqrt{5} - \frac{2^8}{3^4}$. So, you can use tools from calculus to show that this function is positive at x = 2:
$f(x) = \sqrt{x}\sqrt{5} - \frac{x^8}{3^4}$
First of all, its zero is at:
$\frac{x^8}{\sqrt{x}} = 3^4 \sqrt{5} $
This gives us:
$x= 3^{(8/15)} 5^{(1/15)}$
This is slightly larger than 2.
Also, the derivative of f(x) is:
$f'(x) = \frac{\sqrt{5}}{2 \sqrt{x}} - \frac{8x^7}{3^4}$
This is negative for $x > (\frac{3}{2})^{(8/15)} 5^{(1/15)}$ . So, $f(x)$ must be positive before $x= 3^{(8/15)} 5^{(1/15)}$.
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