### Instruction: Real numbers $a$ and $b$ are chosen with $1<a<b$ such that no triangle with positive area has side lengths $1, a,$ and $b$ or $\tfrac{1}{b}, \tfrac{1}{a},$ and $1$. What is the smallest possible value of $b$?### Response: I need to find the smallest $b$ that satisfies the given conditions.
Since no triangle with positive area has side lengths $1, a,$ and $b$, that means either $b \geq 1 + a$ or $a \geq b + 1$.
But the second inequality contradicts $1 < a < b$, so I can eliminate that option.
Therefore, $b \geq 1 + a$.
Similarly, since no triangle with positive area has side lengths $\tfrac{1}{b}, \tfrac{1}{a},$ and $1$, that means either $\tfrac{1}{a} \geq \tfrac{1}{b} + 1$ or $\tfrac{1}{b} \geq \tfrac{1}{a} + 1$.
The first inequality simplifies to $a \leq \tfrac{b}{b + 1}$, and the second one simplifies to $b \leq \tfrac{a}{a + 1}$.
But the second inequality contradicts $1 < a < b$, so I can eliminate that option.
Therefore, $a \leq \tfrac{b}{b + 1}$.
Now I have two inequalities involving $a$ and $b$: $b \geq 1 + a$ and $a \leq \tfrac{b}{b + 1}$.
I want to find the smallest $b$ that satisfies both of them.
I can try to combine them into one inequality by substituting $a \leq \tfrac{b}{b + 1}$ into $b \geq 1 + a$.
This gives me $b \geq 1 + \tfrac{b}{b + 1}$.
I can simplify this by multiplying both sides by $b + 1$ and rearranging the terms.
This gives me $b^2 - b - 1 \geq 0$.
I can solve this quadratic inequality by finding the zeros of $b^2 - b - 1$ and checking the sign of the expression between them.
The zeros are $b = \tfrac{1 \pm \sqrt{5}}{2}$.
The expression is positive for $b \geq \tfrac{1 + \sqrt{5}}{2}$ and negative for $b \leq \tfrac{1 - \sqrt{5}}{2}$.
But since $b > 1$, I can ignore the negative solution.
Therefore, the smallest possible value of $b$ that satisfies both inequalities is $b = \tfrac{1 + \sqrt{5}}{2}$.