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Russia — Regional Round

Algebra Difficulty 4.6 AIME Prove it Russia

Let b>a>1b > a > 1 be real numbers. Define
xn=2n(2nb−2na). x_n = 2^n \left( 2^n \sqrt{b} - 2^n \sqrt{a} \right).
Prove that the sequence x1,x2,…x_1, x_2, \dots is strictly decreasing.

Solution

Докажем, что xn>xn+1x_n > x_{n+1}. Положим A=2n+1aA = 2^{n+1}\sqrt{a} и B=2n+1bB = 2^{n+1}\sqrt{b}. Легко видеть, что B>A>1B > A > 1, отсюда A+B2>1\frac{A+B}{2} > 1. Тогда имеем
xn+1=2n+1(B−A)>0,xn=2n(B2−A2)=2n+1(B−A)A+B2=xn+1⋅A+B2>xn+1,что и требовалось доказать. x_{n+1} = 2^{n+1}(B-A) > 0, \\ x_n = 2^n(B^2 - A^2) = 2^{n+1}(B-A) \frac{A+B}{2} = x_{n+1} \cdot \frac{A+B}{2} > x_{n+1}, \\ \text{что и требовалось доказать.}

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