I have the following exercise:

Let $a_0 > b_0 > 0,$, with $a_0, b_0 \in \mathbb{R}$. We define $a_n$ and $b_n$:

$$a_{n+1} = \frac{a_n+b_n}{2}$$ $$b_{n+1} = \sqrt{a_n b_n}$$

Prove the following:

1. $a_n \ge b_n \forall n$.
2. $a_n$ is decreasing and $b_n$ is increasing.

Question 2 is simple enough, it's question 1 I'm having trouble with. I tried to use induction: I first show that for $n= 0$ the inequality is true, and then, assuming $a_n \ge b_n$, I want to show that $a_{n+1} \ge b_{n+1}$, i.e., $\frac{a_n+b_n}{2} \ge \sqrt{a_n b_n}$

I have tried for a while but haven't managed to get anything. Does anyone have any suggestions?

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It is consequence of the inequality $$\frac{a^2+b^2}{2}\geq ab,\quad\tag{1}$$ (which holds for any $a,b\in\mathbb{R},$ because you can rewrite it as the triviality $(a-b)^2\geq 0.$)
Infact $a_{n+1}=\frac{a_n+b_n}{2}\geq \sqrt{a_nb_n}=b_{n+1}$