Prove that $ \sqrt[100]{1,1}>1,0009$ As I am looking for an "middle/high school" solution I tried to use AM-GM to crack that inequality, but none of my approaches were successful. The inequality is equivalent to
$$ 11 \times 10000^{100}>10 \times 10009^{100}$$ It confuses me how could it be solved by simple estimating RHS by lower numbers since base of RHS is bigger than base of LHS.
 A: Using Bernoulli's Inequality in the form
$$(1 - x)^t \geqslant 1 - xt$$
where $t \geqslant 1$ and $0 \leqslant x \leqslant 1$, we obtain
$$11(10000)^{100} = 11(10009 - 9)^{100} = 11(10009)^{100} \bigg(1 - \dfrac{9}{10009}\bigg)^{100} \geqslant 11(10009)^{100} \bigg(1 - \frac{100\cdot 9}{10009}\bigg).$$
But $$11\cdot \bigg(1 - \frac{100\cdot 9}{10009}\bigg) = \frac{100199}{10009} > 10.$$
Hence, we conclude that
$$11(10000)^{100} > 10(10009)^{100},$$
which as the OP has noted, is equivalent to
$$\sqrt[100]{1.1} > 1.0009.$$
A: You're close. How can we manipulate that into something that we can AM-GM?
Wishful thinking: In AM-GM, the AM side is "some term raised to a power", and the GM side is "product of some terms". It seems like we have the GM side, but not exactly the AM side.
What can we do to force out the AM side? IE How can we make $ 11 \times 10000^{100}$ look like "some term raised to a power"?
Hint / One possible approach:

 Divide both sides by 11. The LHS is now a term raised to some power.


 However, how many terms are we using? On the LHS, with $10000^{100}$, it seems like we want 100 terms. On the RHS, with $\frac{10}{11} \times 10009^{100}$, it seems like we want $1+100=101$ terms.


 Also, the inequality is pretty tight since $\sqrt[100]{1.1} \approx 1.0009536\ldots $ from Wolfram. This suggests that we need to introduce 1 more term, that will make $\frac{10}{11}$ closer to 10009.


 We do so by multiplying by 10000, which makes the LHS $10000^{101}$, still a "term raised to a power".


Applying the hint, we want to show that
$$ 10000^{101} > \frac{ 100000 } { 11} \times 10009^{100}.$$
This is true because:
$\begin{array} { l l l  } \frac{ 100000 } { 11} \times 10009^{100} & < ( \frac{\frac{100000}{11} + 10009 \times 100}{101}) ^ {101} \quad & \text{by AM-GM} \\     
& < 10000 ^ { 101} \quad  & \text{by calculating the base} \end{array}$

Notes:

*

*It's not immediately obvious what could work, so one has to try various approaches.

*EG If we divided throughout by 11 to get LHS = $ 10000^{100}$, are we still able to AM-GM our way out? Note that the power indicates the (weighted) number of terms.

A: When $u\ge0$ we have $u-\frac{u^2}{2}\le\log(1+u)\le  u$, hence
\begin{equation}
\log(1.0009)\le 0.0009< 0.00095 = 0.001 - \frac{0.0001}{2} = 10^{-2}(10^{-1}- \frac{10^{-2}}{2})\le 10^{-2}\log(1.1)
\end{equation}
hence
\begin{equation}
1.0009 < 1.1^{0.01}
\end{equation}
