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 3 of 4: Compress the three-variable inequality into

s=(a+b)/c

Previous step Next step

\frac{a}{b}+\frac{b}{a}\ge 2 \implies \frac{a+b}{c}+\frac{c}{a}+\frac{c}{b}<5;\quad \frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b} \implies \frac{a+b}{c}+\frac{4c}{a+b}<5

Detailed analysis

Since $a/b+b/a\ge 2$ by AM–GM, Step 2 gives $(a+b)/c + c(1/a+1/b) < 5$ . By Cauchy–Schwarz (or AM–HM), $1/a+1/b\ge 4/(a+b)$ , so multiplying by $c>0$ yields $(a+b)/c + 4c/(a+b) < 5$ . Setting $s=(a+b)/c>0$ , the inequality becomes $s + 4/s < 5$ .