A Vietnamese Olympiad Problem: 1968: RMO Training for Algebra

Problem:

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

1. Prove that if m and n are positive integers with $m, then $a^{m}-a^{n} \geq b^{m}-b^{n}>0$.
2. 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.

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: $f_{n}(1)=1-b^{n}-a^{n}=a+b-b^{n}-a^{n}=(a-a^{n})+(b-b^{n}) \geq 0$, $f_{n}(-1)=1+b^{n}-a^{n}=(1-a^{n})+b^{n} \geq 0$, $\frac{S}{2} = \frac{x_{1}+x_{2}}{2}=\frac{b^{n}}{2} \in (-1,1)$.

We conclude that $x_{1}, x_{2} \in [-1, 1]$.

Cheers,

Nalin Pithwa.

Reference: Selected Problems of the Vietnamese Mathematical Olympiad (1962-2009), Le Hai Chau, Le Hai Khoi.