inequality $10<2^{2^{\frac {3}{\log_2 \log_2 10}}}$

Question

While working on this question I ended up with $10<2^2{^{\frac {3}{\log_2 \log_2 10}}}$ I am looking for answers using methods similar to this or this or this or this. Alternative original inequality was $3>(\log_2 \log_2 10)^2$ $$3>(\log_2 \log_2 10)^2$$ $$\frac{3}{(\log_2 \log_2 10)}>(\log_2 \log_2 10)$$ $$2^\frac{3}{(\log_2 \log_2 10)}>2^{(\log_2 \log_2 10)}$$ $$2^\frac{3}{(\log_2 \log_2 10)}>{ \log_2 10}$$ $$2^{2^\frac{3}{(\log_2 \log_2 10)}}>2^{ \log_2 10}$$ $$2^{2^\frac{3}{(\log_2 \log_2 10)}}> 10$$

Answer 1

I made a (stupid) mistake in my previous answer and this one is now in a different spirit.

$$3>(\log_2 \log_2 10)^2 \iff$$

$$\sqrt{3}>\log_2 \log_2 10 \iff$$

$$2^{2^\sqrt{3}}> 10$$

Now, $\sqrt{3}>\frac{989}{571}$.

Also, $$2^{\frac{989}{571}}>\frac{877}{264}$$

Hence $$2^{2^{\sqrt{3}}} > 2^{2^{\frac{989}{571}}} > 2^{\frac{877}{264}} >10$$

This was all done by computer of course, but I have done it so that it is all verifiable by hand (at least to the extent of only needing powers of natural numbers and patience).