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Try this beautiful problem from AMC 12 based on **Triangle inequality problem**.

* 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?*

- \(\frac {3+\sqrt 3}{2}\)
- \(\frac {5}{2}\)
- \(\frac {3+\sqrt 5}{2}\)
- \(\frac {3+\sqrt 6}{2}\)

Triangle Inequality

Inequality

Geometry

But try the problem first...

Answer: \(\frac {3+\sqrt 5}{2}\)

Source

Suggested Reading

American Mathematics Competition - 12B ,2014, Problem Number - 13

Secrets in Inequalities.

First hint

*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 :**

**a+1>b **

**a>b-1**

**\(\frac {1}{a} + \frac {1}{b} >1 \)**

**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.**......................

If you are still stuck after first hint try this ....................

**We already know \(\frac {1}{a} + \frac {1}{b} = 1\)**

**After substituting we will get :**

**\(\frac {1}{b - 1} + \frac {1}{b} = \frac {b+b-1}{b(b-1)} = 1 \)**

**\(\frac {2b - 1}{b(b-1)} = 1 \)**

*Now do the rest of the calculation .............................*

Final Step

*Here is the rest of steps to check your problem :*

**\( 2b - 1 = b^2 - b \)**

**Now Solving for b using the quadratic equation, we get **

**\(b^2 - 3b + 1 = 0 \)**

**\(b = \frac {3 + \sqrt 5}{2} \) (Answer) **

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