Try this beautiful problem from American Mathematics Competition - 12B ,2014, Problem Number - 13 based on Triangle inequality
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 \(\frac {1}{b},\frac {1}{a}\) and 1. What is the smallest possible value of b?
Triangle Inequality
Inequality
Geometry
Answer: \(\frac {3+\sqrt 5}{2}\)
Here is the first hint where you can start this sum:
It is given $1>\frac{1}{a}>\frac{1}{b}$. Use Triangle Inequality here :
$$
\begin{aligned}
& a+1>b \
& a>b-1 \
& \frac{1}{a}+\frac{1}{b}>1
\end{aligned}
$$
If we want to find the lowest possible value of $b$, we create we try to create two degenerate triangles where the sum of the smallest two sides equals the largest side. Thus we get : $a=b-1$
Now try to do the rest of the sum……………………
Here is the rest of steps to check your problem :
$$
2 b-1=b^2-b
$$
Now Solving for $\mathbf{b}$ using the quadratic equation, we get
$$
\begin{aligned}
& b^2-3 b+1=0 \
& b=\frac{3+\sqrt{5}}{2} \text { (Answer) }
\end{aligned}
$$
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