Prove inequality: recurrence sequence

3 The sequence of positive numbers \(u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots\) is such that \(u _ { 1 } > 4\) and, for \(n \geqslant 1\), $$u _ { n + 1 } = \frac { u _ { n } ^ { 2 } + u _ { n } + 12 } { 2 u _ { n } }$$

1. By considering \(\mathrm { u } _ { \mathrm { n } + 1 } - 4\), or otherwise, prove by mathematical induction that \(\mathrm { u } _ { \mathrm { n } } > 4\) for all positive integers \(n\).
2. Show that \(u _ { n + 1 } < u _ { n }\) for \(n \geqslant 1\).

3 The sequence of real numbers \(a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots\) is such that \(a _ { 1 } = 1\) and $$a _ { n + 1 } = \left( a _ { n } + \frac { 1 } { a _ { n } } \right) ^ { 3 }$$

1. Prove by mathematical induction that \(\ln a _ { n } \geqslant 3 ^ { n - 1 } \ln 2\) for all integers \(n \geqslant 2\).
   [0pt] [You may use the fact that \(\ln \left( x + \frac { 1 } { x } \right) > \ln x\) for \(x > 0\).]
2. Show that \(\ln \mathrm { a } _ { \mathrm { n } + 1 } - \ln \mathrm { a } _ { \mathrm { n } } > 3 ^ { \mathrm { n } - 1 } \ln 4\) for \(n \geqslant 2\).

3 The sequence \(x _ { 1 } , x _ { 2 } , x _ { 3 } , \ldots\) is such that \(x _ { 1 } = 3\) and $$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 2 } + 4 x _ { n } - 2 } { 2 x _ { n } + 3 }$$ for \(n = 1,2,3 , \ldots\). Prove by induction that \(x _ { n } > 2\) for all \(n\).

3 The sequence \(a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots\) is such that \(a _ { 1 } > 5\) and \(a _ { n + 1 } = \frac { 4 a _ { n } } { 5 } + \frac { 5 } { a _ { n } }\) for every positive integer \(n\).
Prove by mathematical induction that \(a _ { n } > 5\) for every positive integer \(n\). Prove also that \(a _ { n } > a _ { n + 1 }\) for every positive integer \(n\).

8 The sequence of real numbers \(a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots\) is such that \(a _ { 1 } = 1\) and $$a _ { n + 1 } = \left( a _ { n } + \frac { 1 } { a _ { n } } \right) ^ { \lambda }$$ where \(\lambda\) is a constant greater than 1 . Prove by mathematical induction that, for \(n \geqslant 2\), $$a _ { n } \geqslant 2 ^ { \mathrm { g } ( n ) }$$ where \(g ( n ) = \lambda ^ { n - 1 }\). Prove also that, for \(n \geqslant 2 , \frac { a _ { n + 1 } } { a _ { n } } > 2 ^ { ( \lambda - 1 ) \mathrm { g } ( n ) }\).

1. Express \(\frac{6x + 10}{x + 3}\) in the form \(p + \frac{q}{x + 3}\), where \(p\) and \(q\) are integers to be found. [1]

The sequence of real numbers \(u_1, u_2, u_3, ...\) is such that \(u_1 = 5.2\) and \(u_{n+1} = \frac{6u_n + 10}{u_n + 3}\).

1. Prove by induction that \(u_n > 5\), for \(n \in \mathbb{Z}^+\). [4]