Induction mathematics

Question

Assume $a_1 = 4$ and $a_{n+1} = \sqrt{3+2a_n}$ for all integers $n ≥ 1$. Show with induction that $\forall n ≥ 1, \space a_n > a_{n+1} > 3$. Help me solve this please

Answer 1

hint: $a_n - a_{n+1} = \sqrt{3+2a_{n-1}}-\sqrt{3+2a_n} = \dfrac{2(a_{n-1}-a_n)}{\sqrt{3+2a_{n-1}}+\sqrt{3+2a_n}}$, and $a_{n+1} = \sqrt{3+2a_n} > \sqrt{3+2(3)} = 3$

Answer 2

It clearly holds for $n=1$ since $a_1 = 4$ and $a_2 = \sqrt{11}$.

Assuming it holds for some $k$, that is, $\space a_k > a_{k+1} > 3$, then it holds for $a_{k+1}$ since

$a_{k+2} = \sqrt{3+2a_{k+1}} < a_{k+1} = \sqrt{3+2a_{k}} $ since $a_{k} > a_{k+1}$

and $a_{k+2} = \sqrt{3+2a_{k+1}} > \sqrt{9} = 3$ since $a_{k+1}>3$.

Note that we use the fact that the square-root function is strictly increasing.

Answer 3

You can do it in two parts. First part: prove that $a_n>3$, for every $n$; second part: prove that $a_n>a_{n+1}$, for every $n$.

First part

The induction base is obvious, as $a_1=4>3$. Suppose $a_n>3$. Then $a_{n+1}=\sqrt{3+2a_n}>\sqrt{3+6}=\sqrt{9}=3$.

Second part

Since $a_n>3>0$, the inequality $a_n>a_{n+1}$ is equivalent to $$ a_n^2>3+2a_n $$ that is, $$ a_n^2-2a_n-3>0 $$ or $$ (a_n-3)(a_n+1)>0 $$ which is true, because $a_n>3$.