A Vietnamese Olympiad Problem: 1968: RMO Training for Algebra

Problem:

Let a and b satisfy $a \geq b >0$ and $a+b=1$ .

Prove that if m and n are positive integers with $m<n$ , then $a^{m}-a^{n} \geq b^{m}-b^{n}>0$ .
For each positive integer n, consider a quadratic function: $f_{n}(x)=x^{2}-b^{n}x-a^{n}$ .

Show that $f(x)$ has two roots that are in between -1 and 1.

Solution:

Let $k=n-m \in (0,n)$ . Consider $a^{m}b-ab^{m}=ab(a^{m-1}-b^{m-1})$ with $m-1 \geq 0$ . Since $a \geq b >0$ , we have $a^{m-1} \geq b^{m-1}$ . Hence, $a^{m}b \geq ab^{m}$ . Call this relationship I.

On the other hand, notice that $a \geq b$ , $a^{2} \geq b^{2}$ , $\ldots$ , $a^{k-1} \geq b^{k-1}$ , which implies

$1+a+a^{2}+\ldots+a^{k-1} \geq 1+b+b^{2}+ \ldots + b^{k-1}$ ….call this relationship II.

From relationships I and II, it follows that

$a^{m}b(1+a+a^{2}+\ldots+a^{k-1}) \geq ab^{m}(1+b+b^{2}+\ldots+b^{k-1})$

which can be written as

$a^{m}(1-a)(1+a+a^{2}+\ldots+a^{k-1}) \geq b^{m}(1-b)(1+b+b^{2}+\ldots+b^{k-1})$ , or equivalently, $a^{m}(1-a^{k}) \geq b^{m}(1-b^{k})$ . That is, $a^{m}-a^{n} \geq b^{m}-b^{n}$ .

It remains to prove that $b^{m}-b^{n}>0$ . Indeed, $b^{m}-b^{n}=b^{m}(1-b^{k})>0$ as $0<b<1$ .

The equality occurs if and only if $a=b=\frac{1}{2}$ .

2) Since discriminant $\Delta=b^{2n}+4a^{n}>0$ , $f_{n}(x)$ has two distinct real roots $x_{1} \neq x_{2}$ . Also, note that if $a, b \in (0,1)$ , then the following holds: