A Vietnamese Olympiad Problem: 1968: RMO Training for Algebra


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<n, 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.


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:

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].


Nalin Pithwa.

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

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