## Problem 1

$\mathrm{n}>1$, and $\mathrm{a}$ and $\mathrm{b}$ are positive real numbers such that $\mathrm{a}^{\mathrm{n}}-\mathrm{a}-1=0$ and $\mathrm{b}^{2 \mathrm{n}}-\mathrm{b}-3 \mathrm{a}=0$. Which is larger?


## Solution

Answer: $\mathrm{a}>\mathrm{b}$.

Note that $\mathrm{a}^{\mathrm{n}}=\mathrm{a}+1>1$ (since $\mathrm{a}$ is positive). Hence $\mathrm{a}>1$. So $\mathrm{a}^{2 \mathrm{n}}=(\mathrm{a}+1)^{2}=\mathrm{a}^{2}+2 \mathrm{a}+1$. Put $\mathrm{a}=1+k$, then $a^{2}=1+2 k+k^{2}>1+2 k$, so $a^{2}+1>2+2 k=2 a$. Hence $a^{2 n}>4 a$. So $(b / a)^{2 n}=(b+3 a) / a^{2 n}<(b+3 a) / 4 a$. If $\mathrm{b} / \mathrm{a} \geq 1$, then $(\mathrm{b}+3 \mathrm{a}) / 4 \mathrm{a} \leq(\mathrm{b}+3 \mathrm{~b}) / 4 \mathrm{a}=\mathrm{b} / \mathrm{a}$, so $(\mathrm{b} / \mathrm{a})^{2 \mathrm{n}}<\mathrm{b} / \mathrm{a}$. Contradiction. Hence $\mathrm{b} / \mathrm{a}<1$.

