Prove inequality of big powers without calculating them

Question

How do you prove that $10^{10} > 2 \cdot 9^{10}$ ? I checked with calculator and it's true but I can't prove it without calculating these powers. Use of logarithms also require calculator to actually check that and I can't find other methods that could help me with this.

Answer 1

$$ \begin{align} 10^{10} - 9^{10} & = (10-9)(10^9 + 10^8 \cdot 9 + \;\cdots\; + 10 \cdot 9^8 + 9^9) \\ & \gt 9^9 + 9^8 \cdot 9 + \;\cdots\; + 9 \cdot 9^8 + 9^9 \\ & = 10 \cdot 9^9 \\ & \gt 9 \cdot 9^9 \\ & = 9^{10} \end{align} $$

Answer 2

Use the binomial theorem:

$$\begin{align*} \left(\frac{10}9\right)^{10}&=\left(1+\frac19\right)^{10}\\ &=1+10\cdot\frac19+\binom{10}2\left(\frac19\right)^2+\ldots\\ &>1+10\cdot\frac19\\ &>2\;. \end{align*}$$

Answer 3

Bernoulli's inequality is enough: $$(1+x)^n\ge 1+nx\quad\text{for all}\enspace x>-1.$$ $$\text{Thus}\hspace{5.5em}10^{10}=9^{10}\Bigl(1+\frac19\Bigr)^{10}\ge9^{10}\Bigl(1+\frac{10}9\Bigr)=9^{10}\Bigl(2+\frac19\Bigr)>2\cdot9^{10}.\hspace{5.em}\mbox{}$$