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Lemma
Inequalities Difficulty 4

TMUA practice: Inequalities, problem 1

A TMUA-calibre inequalities problem at difficulty 4 of 5, with the full worked solution.

The question
Real numbers aa and bb are chosen with 1<a<b1<a<b such that no triangle with positive area has side lengths 11 , aa , and bb or 1b\frac{1}{b} , 1a\frac{1}{a} , and 11 . What is the smallest possible value of bb ?
A 3+32\frac{3+\sqrt{3}}{2}
B 52\frac{5}{2}
C 3+52\frac{3+\sqrt{5}}{2}
D 3+62\frac{3+\sqrt{6}}{2}
E 33

The correct answer is highlighted. C

Worked solution

Notice that 1>1a>1b1>\frac{1}{a}>\frac{1}{b} . Using the triangle inequality, we find

a+1>b  ⟹  a>b−1a+1 > b \implies a>b-1

1a+1b>1\frac{1}{a}+\frac{1}{b} > 1

In order for us the find the lowest possible value for bb , we try to create two degenerate triangles where the sum of the smallest two sides equals the largest side. Thus we get

a=b−1a=b-1

and

1a+1b=1\frac{1}{a} + \frac{1}{b}=1

Substituting, we get

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

2b−1b(b−1)=1\frac{2b-1}{b(b-1)} = 1

2b−1=b2−b2b-1=b^2-b

Solving for bb using the quadratic equation, we get

b2−3b+1=0  ⟹  b=C 3+52b^2-3b+1=0 \implies b = \boxed{\textbf{\(C\)} \ \frac{3+\sqrt{5}}{2}}