Let $a_n$ be the sequence defined inductively by $a_1 = 2$ and $a_{n+1} =(1/2)(a_n+2/a_n)$. Prove that $(a_n)^2 ⩾ 2$ for $n ∈ N$.

Question

• I have already shown that $a_{n} \in \left[1, 2\right]\ \forall\, n \in \mathbb{N}$.
• I've also been told to start the proof by assuming $a_{n} < 2$ and finding a contradiction but I have no clue how to do this.

Answer 1

Proof via induction. Basis case: $a_1 > \sqrt{2}.$

Induction case: Using AM-GM, $a_n > 2 \implies a_{n+1} = \frac{a_n + \frac{2}{a_n}}{2} > \sqrt{(a_n)\left( \frac{2}{a_n} \right)} = \sqrt{2}. $

Answer 2

We'll proceed by induction in order to prove that for the sequence defined by $a_{n+1} = \left(a_n + \frac{2}{a_n}\right) \cdot \frac{1}{2}, a_1 = 2$, we have $(a_n)^2 \geq 2$ for all $n$.
We'll first check the base case, and then assume the property holds for $n = k$ to prove it for $n = k+1$.

1-Base Case (n=1)
For $a_1 = 2$, we have: $ (a_1)^2 = 2^2 = 4 \geq 2 $ So, the base case holds.

2- Assumption of the Induction
Assume for some $k \geq 1$, $(a_k)^2 \geq 2$.

3- Inductive Step
Given: $ a_{k+1} = \left(a_k + \frac{2}{a_k}\right) \cdot \frac{1}{2} $ we need to show that $(a_{k+1})^2 \geq 2$.
Let's express $(a_{k+1})^2$ and show it satisfies $(a_{k+1})^2 \geq 2$: $(a_{k+1})^2 = \left(\left(a_k + \frac{2}{a_k}\right) \cdot \frac{1}{2}\right)^2 = \frac{1}{4}\left(a_k^2 + 4 + \frac{4}{a_k^2}\right) = \frac{(a_k^2 + 2)^2}{4a_k^2}$
Now it lets us to prove that $(a_{k+1})^2 \geq 2$. This is a case, indeed the following inequalities are verify:
$\frac{(a_k^2 + 2)^2}{4a_k^2} \geq 2 \Leftrightarrow (a_k^2 + 2)^2 = a_k^4 + 4 + 4a_k^2 \geq 8 a_k^2 \Leftrightarrow a_k^4 + 4 - 4a_k^2 \geq 2^2 + 4 - 4 \cdot 2 =0$
When the last inequality came from the fact that $a_k^2 \geq 2 $ by the assumption of the induction.

Q.E.D.

Answer 3

It's easy to show that the function $f(x)=x+2/x$ has minimum value $2\sqrt2$ for $x>0$; hence $a_{n}$ has minimum value $\sqrt2$ for all $n$

Answer 4

You have $a_{n+1}=\dfrac{a_n^2+2}{2a_n}$ and $a_2=\dfrac32$ so $a_2^2\gt2$.

Consider tha function $f(x)=\dfrac{x^2+2}{2x}$. It is easy to verify that $f'(x)=\dfrac{x^2-2}{2x^2}\gt0$ when $x\gt2$ so it is increasing for $x\gt2$ (in fact for $x\gt\sqrt2$. Since $a_2^2\gt2$ we have $a_n^2\ge2$