Problem 4

Let

n\ge 3

be an integer and

t_1,t_2,\dots,t_n

positive real numbers such that

n^2+1 > (t_1+t_2+\cdots+t_n)\left(\frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_n}\right).

Show that

t_i,t_j,t_k

are the lengths of the sides of a triangle for all

i,j,k

with

1\le i<j<k\le n

Step 1 of 4: Expand into diagonal terms and symmetric pairs

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\left(\sum_{r=1}^n t_r\right)\left(\sum_{s=1}^n \frac{1}{t_s}\right)=n+\sum_{1\le r<s\le n}\left(\frac{t_r}{t_s}+\frac{t_s}{t_r}\right)

Detailed analysis

Fix any three indices $i<j<k$ ; by symmetry we may relabel them as $1,2,3$ and write $(t_1,t_2,t_3)=(a,b,c)$ , ordered so that $a\le b\le c$ . Expanding the product on the right of the given inequality separates the $n$ diagonal terms $t_r/t_r=1$ (which sum to $n$ ) from the $\binom{n}{2}$ unordered pairs $t_r/t_s+t_s/t_r$ with $1\le r<s\le n$ .