Solutions

(a) We use induction to prove that $x_{n}^2 \geq 2$ for all $n \in \mathbb{N}$. The base case $x_1^2 = 4 \geq 2$ holds. Now, assuming that $x_{n}^2 \geq 2$, we prove the induction step:

$$\begin{aligned} x_{n + 1}^2 &= \frac{1}{4} (x_{n} + \frac{2}{x_{n}})^2 \\ &= \frac{1}{4} (x_{n}^2 + 4 + \frac{4}{x_{n}^2}) \\ &= 1 + \frac{x_{n}^2}{4} + \frac{1}{x_{n}^2}. \end{aligned}$$

Notice now that

$$\begin{aligned} x_{n + 1}^2 = 1 + \frac{x_{n}^2}{4} + \frac{1}{x_{n}^2} \geq 2 &\iff \frac{x_{n}^2}{4} + \frac{1}{x_{n}^2} \geq 1 \\ &\iff \frac{x_{n}^{4} + 4 }{4x_{n}^2} \geq 1 \\ &\iff x_{n}^{4} + 4 \geq 4x_{n}^2 \\ &\iff x_{n}^{4} - 4x_{n}^2 + 4 \geq 0 \\ &\iff {(x_{n}^2 - 2)}^{2} \geq 0 \end{aligned}$$

with the last inequality being obviously true since the square of any real number is greater or equal to 0.
Now, we prove that the sequence is decreasing. To assist us in proving that the sequence is decreasing, let us first prove that the sequence is strictly positive.
As always, we use induction. The base case holds since

$$x_1 = 2 > 0.$$

The induction step also is quite easy. Assume that $x_{n} > 0$, then

$$\begin{aligned} x_{n + 1} &= \frac{1}{2} (x_{n} + \frac{2}{x_{n}}) \\ &= \frac{1}{2}x_{n} + \frac{1}{x_{n}} \\ &= \frac{x_{n} + 2}{2x_{n}} > 0 \end{aligned}$$

since both the numerator and the denominator of the fraction are positive.
We now move on to use induction to prove that the sequence is decreasing. The base case holds because

$$x_1 = 2 \geq \frac{3}{2} = x_2.$$

For the induction step, assume that $x_{n}^2 \geq 2$, which we've previously proved to be true (this is where we make use of the inequality, which the exercise asks us to), then

$$\begin{aligned} x_{n} - x_{n + 1} &= x_{n} - \frac{1}{2} (x_{n} + \frac{2}{x_{n}}) \\ &= x_{n} - \frac{1}{2}x_{n} + \frac{1}{x_{n}} \\ &= \frac{1}{2}x_{n} + \frac{1}{x_{n}} \\ &= \frac{x_{n}^2 + 2}{2x_{n}} \geq 0. \end{aligned}$$

Thus, by MCT the sequence $x_{n}$ converges and, by denoting $\lim_{n \to \infty} x_{n}$ as $L$, the limit is

$$\begin{aligned} \lim_{n \to \infty} x_{n} = \lim_{n \to \infty} \frac{1}{2} ( x_{n} + \frac{2}{x_{n}} ) &\implies L = \frac{1}{2} (L + \frac{2}{L}) \\ &\implies L = \frac{L}{2} + \frac{1}{L} \\ &\implies \frac{L}{2} = \frac{1}{L} \\ &\implies L^2 = 2 \\ &\implies L = \pm \sqrt{2}. \end{aligned}$$

Since we've proven before that the sequence is strictly positive, then the limit of $x_{n}$ cannot be negative by the Order Limit Theorem. Therefore the limit of $x_{n}$ is $\sqrt{2}$.

(b) Notice that the sequence

$$x_{n + 1} = \frac{1}{2} (x_{n} + \frac{c}{x_{n}})$$

converges to $\sqrt{c}$ by setting an initial value of the sequence greater or equal to $c$.