a)$ (x_1^n + x_2^n)(x_1^{n + 2} + x_2^{n + 2}) > (x_1^{n + 1} + x_2^{n + 1})^2$. $ \Rightarrow \frac {S_{n + 1}}{S_{n + 2}} < \frac {S_n}{S_{n + 1}}$. b)$ Lim_{n\rightarrow + \infty}{\frac {S_{n}}{S_{n + 1}} = \frac {a + \sqrt {a^2 - 4}}{2}}$. So $ Lim_{n\rightarrow + \infty}{\frac {\sum_{i = 1}^n{\frac {S_i}{S_{i + 1}}}}{n} = \frac {a + \sqrt {a^2 - 4}}{2}}$. $ \Rightarrow \frac {a + \sqrt {a^2 - 4}}{2}\geq 1$. $ \Rightarrow a \geq 2$. *) If $ a\geq 2$, then $ \sum_{i = 1}^n{\frac {S_i}{S_{i + 1}}} > n.\frac {a + \sqrt {a^2 - 4}}{2}\geq n > n - 1$.

a) $ x_1,\ x_2$ are real solutions of $ x^2 + 1 = ax$ $ \Longleftrightarrow x_1,\ x_2$ are the $ x$ coordinates of the parabola $ C: y = x^2 + 1$ and the line $ l: y = ax$. For $ a\geq 2$, When $ a = 2$, $ C$ touches $ l$ at $ x = 2$, when $ a > 2$, $ C$ intersects with $ l$ at two points, so $ x_1,\ x_2$ are positive real numbers. $ \therefore x_1^2 = ax_1 - 1,\ x_2^2 = ax_2 - 1\Longrightarrow x_1^{n + 2} = ax_1^{n + 1} - x_1^n,\ x_2^{n + 2} = ax_2^{n + 1} - x_2^n$. Adding both sides gives $ \therefore S_{n + 2} = aS_{n + 1} - S_n$. Thus we have $ S_{n + 1}^2 - S_nS_{n + 2} = (x_1^{n + 1} + x_2^{n + 1})^2 - (x_1^n + x_2^n)(x_1^{n + 2} + x_2^{n + 2}) = - x_1^n x_2^n(x_1 - x_2)^2\leq 0$. Since $ S_n > 0$, $ \therefore \frac {S_{n}}{S_{n + 1}}\geq \frac {S_{n + 1}}{S_{n + 2}}$ Comment 1: By Vieta, $ x_1 + x_2 = a\geq 2,\ x_1x_2 = 1,\ D = a^2 - 4\geq 0$ gives $ x_1 > 0,\ x_2 > 0$. Comment 2: $ x = 0$ is not solution of $ x^2 - ax + 1 = 0$, so we have $ x + \frac {1}{x} = a$, since $ x\cdot \frac {1}{x} = 1 > 0$, $ x$ and $ \frac {1}{x}$ have same sign, by $ a\geq 2$, we get $ x > 0$. b) This is Wrong. $ S_{n + 2} = aS_{n + 1} - S_n\ (S_n > 0)\ \Longleftrightarrow \frac {S_{n + 2}}{S_{n + 1}} = a - \frac {S_n}{S_{n + 1}}$ $ \Longleftrightarrow \frac {S_{n + 2}}{S_{n + 1}} - \frac {a}{2} = - \left(\frac {S_n}{S_{n + 1}} - \frac {a}{2}\right)$, yielding $ \frac {S_{n + 1}}{S_n} - \frac {a}{2} = ( - 1)^{n - 1}\left(\frac {S_1}{S_2} - \frac {a}{2}\right)$ $ \therefore \sum_{k = 1}^n \frac {S_k}{S_{k + 1}} = \sum_{k = 1}^n \left\{\frac {a}{2} + ( - 1)^{k - 1}\left(\frac {S_1}{S_2} - \frac {a}{2}\right)\right\}$ $ = \frac {a}{2}n + \left(\frac {S_1}{S_2} - \frac {a}{2}\right)\cdot \frac {1 - ( - 1)^n}{1 - ( - 1)}$ When $ n$ is even, $ \sum_{k = 1}^n \frac {S_k}{S_{k + 1}} = \frac {a}{2}n - 1 > n - 1\ \because a\geq 2$. When $ n$ is odd, $ \sum_{k = 1}^n \frac {S_k}{S_{k + 1}} = \frac {a}{2}n + \frac {S_1}{S_2} - \frac {a}{2} = \frac {a}{2}(n - 1) + \frac {S_1}{S_2}\geq n - 1 + \frac {S_1}{S_2} > n - 1$.

kunny wrote: a) $ x_1,\ x_2$ are real solutions of $ x^2 + 1 = ax$ $ \Longleftrightarrow x_1,\ x_2$ are the $ x$ coordinates of the parabola $ C: y = x^2 + 1$ and the line $ l: y = ax$. For $ a\geq 2$, When $ a = 2$, $ C$ touches $ l$ at $ x = 2$, when $ a > 2$, $ C$ intersects with $ l$ at two points, so $ x_1,\ x_2$ are positive real numbers. $ \therefore x_1^2 = ax_1 - 1,\ x_2^2 = ax_2 - 1\Longrightarrow x_1^{n + 2} = ax_1^{n + 1} - x_1^n,\ x_2^{n + 2} = ax_2^{n + 1} - x_2^n$. Adding bothe sides gives $ \therefore S_{n + 2} = aS_{n + 1} - S_n$. Thus we have $ S_{n + 1}^2 - S_nS_{n + 2} = (x_1^{n + 1} + x_2^{n + 1})^2 - (x_1^n + x_2^n)(x_1^{n + 2} + x_2^{n + 2}) = - x_1^n x_2^n(x_1 - x_2)^2\leq 0$. Since $ S_n > 0$, $ \therefore \frac {S_{n}}{S_{n + 1}}\geq \frac {S_{n + 1}}{S_{n + 2}}$ Comment 1: By Vieta, $ x_1 + x_2 = a\geq 2,\ x_1x_2 = 1,\ D = a^2 - 4\geq 0$ gives $ x_1 > 0,\ x_2 > 0$. Comment 2: $ x = 0$ is not solution of $ x^2 - ax + 1 = 0$, so we have $ x + \frac {1}{x} = a$, since $ x\cdot \frac {1}{x} = 1 > 0$, $ x$ and $ \frac {1}{x}$ have same sign, by $ a\geq 2$, we get $ x > 0$. b) $ S_{n + 2} = aS_{n + 1} - S_n\ (S_n > 0)\ \Longleftrightarrow \frac {S_{n + 2}}{S_{n + 1}} = a - \frac {S_n}{S_{n + 1}}$ $ \Longleftrightarrow \frac {S_{n + 2}}{S_{n + 1}} - \frac {a}{2} = - \left(\frac {S_n}{S_{n + 1}} - \frac {a}{2}\right)$, yielding $ \frac {S_{n + 1}}{S_n} - \frac {a}{2} = ( - 1)^{n - 1}\left(\frac {S_1}{S_2} - \frac {a}{2}\right)$ $ \therefore \sum_{k = 1}^n \frac {S_k}{S_{k + 1}} = \sum_{k = 1}^n \left\{\frac {a}{2} + ( - 1)^{k - 1}\left(\frac {S_1}{S_2} - \frac {a}{2}\right)\right\}$ $ = \frac {a}{2}n + \left(\frac {S_1}{S_2} - \frac {a}{2}\right)\cdot \frac {1 - ( - 1)^n}{1 - ( - 1)}$ When $ n$ is even, $ \sum_{k = 1}^n \frac {S_k}{S_{k + 1}} = \frac {a}{2}n - 1 > n - 1\ \because a\geq 2$. When $ n$ is odd, $ \sum_{k = 1}^n \frac {S_k}{S_{k + 1}} = \frac {a}{2}n + \frac {S_1}{S_2} - \frac {a}{2} = \frac {a}{2}(n - 1) + \frac {S_1}{S_2}\geq n - 1 + \frac {S_1}{S_2} > n - 1$. Hmmm, can u tell me if i'm right ... i proved that Sn/Sn+1>Sn+k/Sn+k+1 ... for all k E N ... is it right, just for (a)

ridgers wrote: Let $ a\geq 2$ be a real number; with the roots $ x_{1}$ and $ x_{2}$ of the equation $ x^2 - ax + 1 = 0$ we build the sequence with $ S_{n} = x_{1}^n + x_{2}^n$. a)Prove that the sequence $ \frac {S_{n}}{S_{n + 1}}$, where $ n$ takes value from $ 1$ up to infinity, is strictly non increasing. b)Find all value of $ a$ for the which this inequality hold for all natural values of $ n$ $ \frac {S_{1}}{S_{2}} + \cdots + \frac {S_{n}}{S_{n + 1}} > n - 1$ a) $ \frac {S_n}{S_{n + 1}} = \frac {x_1^n + x_2^n}{x_1^{n + 1} + x_2^{n + 1}} = x_1 + \frac {x_1^n - x_1^{n + 2}}{x_1^{n + 1} + x_2^{n + 1}} = x_1 + \frac {1 - x_1^2}{x_1 + x_2^{2n + 1}}$. (Here i use $ x_1x_2 = 1$)Wlog assume that $ x_2 \ge 1 \Rightarrow x_1 \le 1$, and it is obvious that it is non-increasing. b) Assume that $ \frac {S_0}{S_1} = c < 1$. Then by a): $ n - 1 < \frac {S_{1}}{S_{2}} + \cdots + \frac {S_{n}}{S_{n + 1}} \le n \cdot c \iff n < \frac {1}{1 - c}$ which can't hold. So $ \frac {2}{a} = \frac {S_0}{S_1} \ge 1 \iff a \le 2$ which gives $ a = 2$ as the only solution. The ineq clearly holds when $ a = 2$ since $ S_n = 1 \forall n$. @kunny, Thjch Ph4 Trjnh: The conclusion $ a \ge 2$ is wrong. $ a=2$ is correct. For a counterexample try $ a=100,n=2$. Then we have $ S_1 = 100, S_2 = 9998, S_3 = 999970$ and $ \frac{S_1}{S_2} + \frac{S_2}{S_3} < 1 = 2-1$

a) \[ \frac {S_n}{S_{n + 1}}\geq \frac {S_{n + 1}}{S_{n + 2}}\] \[ \iff \frac {x_1 ^n + x_2 ^n}{x_1 ^{n + 1} + x_2 ^{n + 1}}\geq \frac {x_1 ^{n + 1} + x_2 ^{n + 1}}{x_1 ^{n + 2} + x_2 ^{n + 2}}\] \[ \iff x_1^{2n + 2} + x_2 ^{2n + 2} + x_1 ^n x_2 ^n (x_1 ^2 + x_2 ^2) \geq x_1 ^{2n + 2} + 2x_1 ^{n + 1}x_2^{n + 1} + x_2 ^{2n + 2}\] \[ \iff ( x_1 - x_2 )^2 x_1 ^n x_2 ^n \geq 0\] This is true because $ x_1 x_2 = 1 > 0$ b) (Solution Deleted because it was wrong) P.S- Nice ideas kunny and Mathias

enndb0x wrote: For $ n = 2$ we have: \[ \frac {S_1}{S_2} \geq 1 \iff x_1 + x_2 \geq x_1^2 + x_2 ^2\] $ n=2$ is $ \frac{S_1}{S_2} + \frac{S_2}{S_3} > 1$ and not $ \frac{S_1}{S_2} \ge 1$.

Ups, sorry. That affects my proof drastically. There is still a hope that examiners would not see the mistake in part b)

Quote: b)$ Lim_{n\rightarrow + \infty}{\frac {S_{n}}{S_{n + 1}} = \frac {a + \sqrt {a^2 - 4}}{2}}$. is wrong. $ Lim_{n\rightarrow + \infty}{\frac {S_{n}}{S_{n + 1}} = \frac {a - \sqrt {a^2 - 4}}{2}}$. So $ \frac {a - \sqrt {a^2 - 4}}{2}\geq 1$. $ \Rightarrow a=2$.

In my previous post, solution to b) was wrong. Here is the corrected version: a) $ x_1,\ x_2$ are real solutions of $ x^2 + 1 = ax$ $ \Longleftrightarrow x_1,\ x_2$ are the $ x$ coordinates of the parabola $ C: y = x^2 + 1$ and the line $ l: y = ax$. For $ a\geq 2$, When $ a = 2$, $ C$ touches $ l$ at $ x = 2$, when $ a > 2$, $ C$ intersects with $ l$ at two points, so $ x_1,\ x_2$ are positive real numbers. $ \therefore x_1^2 = ax_1 - 1,\ x_2^2 = ax_2 - 1\Longrightarrow x_1^{n + 2} = ax_1^{n + 1} - x_1^n,\ x_2^{n + 2} = ax_2^{n + 1} - x_2^n$. Adding both sides gives $ \therefore S_{n + 2} = aS_{n + 1} - S_n$. Thus we have $ S_{n + 1}^2 - S_nS_{n + 2} = (x_1^{n + 1} + x_2^{n + 1})^2 - (x_1^n + x_2^n)(x_1^{n + 2} + x_2^{n + 2}) = - x_1^n x_2^n(x_1 - x_2)^2\leq 0$. Since $ S_n > 0$, $ \therefore \frac {S_{n}}{S_{n + 1}}\geq \frac {S_{n + 1}}{S_{n + 2}}$ Comment 1: By Vieta, $ x_1 + x_2 = a\geq 2,\ x_1x_2 = 1,\ D = a^2 - 4\geq 0$ gives $ x_1 > 0,\ x_2 > 0$. Comment 2: $ x = 0$ is not solution of $ x^2 - ax + 1 = 0$, so we have $ x + \frac {1}{x} = a$, since $ x\cdot \frac {1}{x} = 1 > 0$, $ x$ and $ \frac {1}{x}$ have same sign, by $ a\geq 2$, we get $ x > 0$. b) From result of a), $ S_{n + 1}^2 - S_nS_{n + 2}$ $ = S_{n + 1}^2 - S_n(aS_{n + 1} - S_n)$ $ = S_n^2 - aS_nS_{n + 1} + S_{n + 1}^2$ $ \therefore \frac {S_n}{S_{n + 1}}\geq \frac {S_{n + 1}}{S_{n + 2}}\ (S_n > 0)$ $ \Longleftrightarrow S_{n + 1}^2 - S_nS_{n + 2}\leq 0$ $ \Longleftrightarrow \left(\frac {S_n}{S_{n + 1}}\right)^2 - a\frac {S_n}{S_{n + 1}} + 1\leq 0$ Since $ \frac {S_n}{S_{n + 1}} > 0$, we have $ \boxed{\frac {a - \sqrt {a^2 - 4}}{2}\leq \frac {S_n}{S_{n + 1}}\leq \frac {a + \sqrt {a^2 - 4}}{2}}$ $ \therefore \sum_{k = 1}^n \frac {S_k}{S_{k + 1}}\geq \frac {a - \sqrt {a^2 - 4}}{2}n = \frac {2}{a + \sqrt {a^2 - 4}}n$. I'm stuck here.

a) We need to prove that $\frac {S_n} {S_{n+1}} < \frac {S_{n-1}} {S_n}$ (for $a=2$ the sequence is constant); after plugging in $s_k=x_1 ^k+x_2^k$ and simplifying everything, it becomes $2x_1x_2<x_1^2+x_2^2$, which rearranges to $(x_1-x_2)^2>0$. b) We want to prove that if $a>2$ then the sum behaves like a multiple of $n$ with coefficient smaller than $1$. Indeed, one of the roots of the equation, say $x_1$, has value larger than $1$, while the other is smaller than $1$. Then simple calculations show that $S_n = x_1^n+O(x_2^n)$, $a_n=\frac 1 {x_1}+o(1)$ and the inequality becomes $\frac n {x_1}+o(n)>n-1$, which is false for $n$ large enough. Then the only viable value is $a=2$, which works.

Cassius wrote: a) We need to prove that $\frac {S_n} {S_{n+1}} < \frac {S_{n-1}} {S_n}$ (for $a=2$ the sequence is constant); after plugging in $s_k=x_1 ^k+x_2^k$ and simplifying everything, it becomes $2x_1x_2<x_1^2+x_2^2$, which rearranges to $(x_1-x_2)^2>0$. b) We want to prove that if $a>2$ then the sum behaves like a multiple of $n$ with coefficient smaller than $1$. Indeed, one of the roots of the equation, say $x_1$, has value larger than $1$, while the other is smaller than $1$. Then simple calculations show that $S_n = x_1^n+O(x_2^n)$, $a_n=\frac 1 {x_1}+o(1)$ and the inequality becomes $\frac n {x_1}+o(n)>n-1$, which is false for $n$ large enough. Then the only viable value is $a=2$, which works. I think a>=2. I made this solution: n((2^(n+1)*a)/((a+(a^2-4)^(1/2))^(n+1)+(a-(a^2-4)^(1/2))^(n+1))+1>0 and also its true in all cases of (a) i think