Showing that a sequence is decreasing I need to show that the sequence defined by $a_1 = \frac{5}{2} $ and $a_{n+1} = \frac{1}{5}(a_n^2 + 6) $
is decreasing. The fact that $2 < a_n < 3 $ has been proven already.
I now just need to show that $a_n$ is decreasing.
I have tried the usual ways of showing $a_{n+1} - a_n \le 0 $ and $\frac{a_{n+1}}{a_n} \le 1 $
but could not get anywhere with these.
 A: Hint
Define $$e_n=a_n-2$$which leads to
$$e_{n+1}=a_{n+1}-2={a_n^2-4\over 5}={e_n}{e_n+4\over 5}$$and prove $e_n$ is decreasing to $0$.
Remark
The reason I subtracted $2$ from $a_n$ is that the sequence seems to tend to $2$, hence this is a good starting point.
A: If $2<a_n<3$, then $(a_n-2)(a_n-3)<0$, so $a_n^2-5a_n+6<0$, so $a_{n+1}=\frac15(a_n^2+6)<a_n.$
A: Using the definition of a decreasing sequence, we starter by computing $a_{n+1}-a_n.$
$$\begin{align}
a_{n+1}-a_n & \implies \frac{1}{5}(a_n^2+6)-a_n\\
& \implies \frac{a_n^2-5a_n+6}{5}\\
& \implies \frac{(a_n - \frac{5}{2})^2-\frac{1}{4}}{5}\\
& \implies \frac{(\frac{2a_n-5}{2})^2-\frac{1}{4}}{5}\\
& \implies \frac{(2a_n^2-5)-1}{20}
\end{align}$$
Next, since you have proved that $2 < a_n < 3,$ observe that
$$\begin{align}
2 < a_n < 3 & \implies 4 < 2a_n < 6\\
& \implies -1 < 2a_n -5 < 1\\
& \implies (2a_n-5)^2<1\\
& \implies (2a_n-5)^2-1<0\\
& \implies \frac{(2a_n-5)^2-1}{20}<0
\end{align}$$
Therefore you have that
$$a_{n+1}-a_n < 0$$
and by definition you conclude that $a_n$ is a decreasing sequence. $\square$
A: If $x\in (2,3)$, then
$$
x^2-5x+6=(x-2)(x-3)<0 \quad\Longrightarrow\quad \frac{1}{5}(x^2+6)<x
$$
So, if $a_n\in(2,3)$, then
$$
a_n > \frac{1}{5}(a_n^2+6)=a_{n+1}
$$
