Which number is greater, $11^{11}$ or $9^{12}$?

$$ \frac{11^{11}}{9^{11}} =\left(\frac{2}{9}+1\right)^{11} = \sum_{k=0}^{11} \binom{11}{k} \left(\frac{2}{9}\right)^k > \sum_{k=0}^5 \binom{11}{k} \left(\frac{2}{9}\right)^k = \frac{177665}{19683} > 9 $$

This is a variant on Servaes's answer. Note that $3^5=243=2\cdot11^2+1$. Using a binomial expansion and some extremely crude upper bounds, we find

$$\begin{align} 3\cdot9^{12} &=3^{25}\\ &=(2\cdot11^2+1)^5\\ &=32\cdot11^{10}+80\cdot11^8+80\cdot11^6+40\cdot11^4+10\cdot11^2+1\\ &\lt32\cdot11^{10}+80\cdot11^8+11^8+11^8+11^8+11^8\\ &\lt32\cdot11^{10}+121\cdot11^8\\ &=32\cdot11^{10}+11^{10}\\ &=3\cdot11^{11} \end{align}$$

and thus $9^{12}\lt11^{11}$.

The goal is to prove that $(1 + 2/9)^{11} > 9$. As the left-hand side is approximately $9.091843$ this will be a bit tricky.

The big idea in this solution is to try to exploit the fact that $(11/9)^2 = 121/81$ is just under $3/2$, since $3/2$ will be simple to work with.

Start with the inequality $3^2 \times 29 > 2^8$, i. e. $261 > 256$. Multiply throuhg on both sides by $3^4 2^3$ to get $3^6 \times 232 > 81 \times 2^{11}$.

Now, we can rewrite this as

$$ 3^{11} \times {232 \over 243} > 81 \times 2^{11}. $$

We then have $(232/243) = 1-11/243 < (1-1/243)^{11}$ (by Bernoulli's inequality, as pointed out by roby5) and so it follows that

$$ 3^{11} \times (1-1/243)^{11} > 81 \times 2^{11}. $$

At this point most of the work is done. Divide both sides by $2^{11}$ to get

$$ 1.5^{11} \times (1-1/243)^{11} > 81 $$

and multiply both sides by $81^{11}$ to get

$$ 121.5^{11} \times (1-1/243)^{11} > 81^{12} $$

But since both factors on the left-hand side are eleventh powers, we can rewrite this as

$$ 121^{11} > 81^{12} $$

and taking square roots of both sides gives the desired result.