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 4 of 4: Factor the quadratic in

s

to get

a+b>c

Previous step Next step

s+\frac{4}{s}<5 \iff s^2-5s+4<0 \iff (s-1)(s-4)<0 \implies s>1 \implies a+b>c

Detailed analysis

Since $s>0$ , multiplying $s+4/s<5$ through by $s$ gives $s^2-5s+4<0$ , which factors as $(s-1)(s-4)<0$ . Thus $1<s<4$ , and in particular $s>1$ , i.e. $a+b>c$ . Because we ordered $a\le b\le c$ , the other two triangle inequalities $a+c>b$ and $b+c>a$ hold automatically, so $a,b,c$ are the sides of a triangle.