Inequality from a Sequence I am working on a problem and I am lead to prove the following inequality which is true based on writing out the sequence and on the fact that it should be true based on what I am trying to prove. 
Let $s_0,s_1,s_2,....$ be a sequence of positive numbers satisfying $s_0=s_1=1$ and $s_n = s_{n-1} - \alpha s_{n-2}$ where $0<\alpha<\tfrac{1}{4}$. Prove that $2s_{n+1}\geq s_n$. 
Perhaps, the following property will help, $s_{n-1}s_{n+1}\leq s_n^2$ that you can assume. 

Letting $A\lt B$ be the roots of the equation $x^2-x+\alpha=0,$ we have
$$s_{n+1}-As_{n}=B(s_{n}-As_{n-1})=\cdots=B^n(s_1-As_0)=B^n(1-A).$$
$$s_{n+1}-Bs_{n}=A(s_{n}-Bs_{n-1})=\cdots=A^n(s_1-Bs_0)=A^n(1-B).$$
Hence, we have
$$s_n=\frac{B^n(1-A)-A^n(1-B)}{B-A}.$$
Hence,
$$2s_{n+1}-s_n\ge 0$$
$$\iff 2\cdot B^{n+1}(1-A)-2\cdot A^{n+1}(1-B)-B^n(1-A)+A^n(1-B)\ge 0$$
$$\iff B^n(2B-1)(A+1)+A^n(1-2A)(1-B)\ge0$$
This is true because of 
$$0\lt A=\frac{1-\sqrt{1-4\alpha}}{2}\lt\frac 12\lt B=\frac{1+\sqrt{1-4\alpha}}{2}\lt 1.$$
Hence, we now know that we prove that $2s_{n+1}-s_n\ge 0$.
