Problem: Let $n \ge 3$ be an integer
and $t_1$, $t_2$, \dots, $t_n$ positive real numbers such that
\[ n^2+1 > \left(t_1 + t_2 + \dots + t_n\right)
  \left( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} \right). \]
Show that $t_i$, $t_j$, $t_k$ are the sides of a triangle
for all $i$, $j$, $k$ with $1 \le i < j < k \le n$.

Solution: Let $a = t_1$, $b = t_2$, $c = t_3$.
Expand:
\begin{align*}
  n^2+1 &> \left(t_1 + t_2 + \dots + t_n\right)
    \left( \frac{1}{t_1} + \dots + \frac{1}{t_n} \right) \\
  &= n + \sum_{1 \le i < j \le n}
    \left( \frac{t_i}{t_j} + \frac{t_j}{t_i} \right) \\
  &= n + \sum_{1 \le i < j \le n}
    \left( \frac{t_i}{t_j} + \frac{t_j}{t_i} \right) \\
  &\ge n + \sum_{1 \le i < j \le 3}
    \left( \frac{t_i}{t_j} + \frac{t_j}{t_i} \right)
    + \sum_{\substack{1 \le i < j \le n \\ j > 3 }} 2 \\
  &= n + 2\left( \binom n2-3 \right)
    + \left( \frac ab + \frac ba \right)
    + \frac{a+b}{c} + \frac{c}{b} + \frac{c}{a} \\
  &\ge n + 2\left( \binom n2-3 \right) + 2
    + \frac{a+b}{c} + c \cdot \frac{4}{a+b}
\end{align*}
So, we conclude that
\[ \frac{a+b}{c} + \frac{4c}{a+b} < 5 \]
which rearranges to
\[ \left( 4c-(a+b) \right)\left( c-(a+b) \right) < 0. \]
This is enough to imply $c < a+b$.

\begin{remark*}
  A variant of the same argument allows one to improve
  the left-hand side to $(n+\sqrt{10}-3)^2$.
  One does so by writing
  \[ \text{RHS} \ge \left( \sqrt{\left( a+b+c \right)
    \left( \frac1a+\frac1b+\frac1c \right)} + (n-3) \right)^2 \]
  and estimating the square root as in the previous approach.

  In addition, $(n+\sqrt{10}-3)^2$ is best possible,
  as seen by taking $(a,b,c) = (2,1,1)$ and $t_4 = t_5 = \dots = \frac25 \sqrt{10}$.
\end{remark*}