2. Find three real numbers $a<b<c$ satisfying:
$$
\begin{aligned}
a+b+c & =21 / 4 \\
1 / a+1 / b+1 / c & =21 / 4 \\
a b c & =1
\end{aligned}
$$
Solution: $1 / 4,1,4$
By inspection, one notices that if $b$ is a number such that $b+1 / b=17 / 4$, then $a=1, c=$ $1 / b$ will work. Again by inspection (or by solving the quadratic $b^{2}-17 b / 4+1=0$ ), one finds $b=1 / 4$ or 4 , so the numbers are $1 / 4,1$, and 4 .
Alternative Solution: Note that $a b+b c+c a=a b c(1 / a+1 / b+1 / c)=21 / 4$, so $a, b$, and $c$ are the roots of the polynomial $x^{3}-21 x^{2} / 4+21 x / 4-1$, which factors as $\frac{1}{4}(x-1)(x-4)(4 x-1)$, giving the same answer.