Solutions

(a) To prove that the inequality holds under the given assumptions, let us first prove another one. That is, we first prove that

$$\forall a,b \in \mathbb{R}_{\geq 0 }, \quad a \leq b \implies a^2 \leq b^2.$$

Since the product of positive numbers is positive, we then have that

$$\begin{aligned} \forall a, b \in \mathbb{R}_{\geq 0}, \quad a \leq b &\implies b - a \geq 0 \\ &\implies (b - a)(b + a) \geq 0 \\ &\implies b^2 - a^2 \geq 0 \\ &\implies a^2 \leq b^2. \end{aligned}$$

Now, coming back to the inequality of the exercise, we have to prove that

$$\begin{aligned} \forall x,y \in \mathbb{R}_{\geq 0}, \quad \sqrt{xy} \leq \frac{(x + y)}{2}. \end{aligned}$$

Applying our freshly proved inequality, we know that

$$\begin{aligned} \forall x,y \in \mathbb{R}_{\geq 0}, \quad \sqrt{xy} \leq \frac{(x + y)}{2} &\iff xy \leq \frac{(x + y)^2}{4} \\ &\iff xy \leq \frac{x^2}{4} + \frac{xy}{2} + \frac{y^2}{4} \\ &\iff \frac{xy}{2} \leq \frac{x^2}{4} + \frac{y^2}{4} \\ &\iff 2xy \leq x^2 + y^2 \\ &\iff 0 \leq x^2 - 2xy + y^2 \\ &\iff 0 \leq (x - y)^2. \end{aligned}$$

Thus, the inequality holds when $x$ and $y$ are positive real numbers.

(b) We first prove that both sequences are convergent. To assist us in proving this, let us first prove that both sequences are positive. That is,

$$\forall n \in \mathbb{N}, \quad x_{n}, y_{n} \geq 0.$$

Using induction, we first prove the base case. Since

$$0 \leq x_1 \leq y_1$$

then

$$x_1, y_1 \geq 0.$$

Moving on to the induction step, assume that both $x_{n}$ and $y_{n}$ are positive. Then, for $x_{n}$, we have

$$x_{n + 1} = \sqrt{x_{n} y_{n}} =\sqrt{x_{n}} \sqrt{y_{n}} \geq 0$$

since each argument of the square root functions are positive. With the same assumption (both $x_{n}$ and $y_{n}$ are positive), for $y_{n}$, we have

$$y_{n + 1} = \frac{(x_{n} + y_{n})}{2} \geq 0$$

since the numerator is obviously positive.
We can now come back to our original objective which was to prove that both sequences are convergent. To apply MCT, let's prove that $x_{n}$ is increasing and bounded above. The base case holds true since

$$\begin{aligned} x_1 \leq y_1 &\implies \sqrt{x_1} \leq \sqrt{y_1} \\ &\implies x_1 \leq \sqrt{x_1} \sqrt{y_1} = x_2. \end{aligned}$$

Next, let's assume that $x_{n + 1} \geq x_{n}$. Then,

$$\begin{aligned} x_{n + 2} \geq x_{n + 1} &\iff \sqrt{x_{n + 1} y_{n + 1}} \geq x_{n + 1} \\ &\iff x_{n + 1}y_{n + 1} \geq x_{n + 1}^2 \\ &\iff y_{n + 1} \geq x_{n + 1}. \end{aligned}$$

The last inequality is obviously true, applying the identity in (a), since both $x_{n}$ and $y_{n}$ are positive. To help us in finding an upper bound for $x_{n}$, let's prove that prove with induction that $y_{n}$ is decreasing. Our base case holds because

$$\begin{aligned} y_2 = \frac{(x_1 + y_1)}{2} \leq y_1 &\iff x_1 + y_1 \leq 2y_1 \\ &\iff x_1 \leq y_1. \end{aligned}$$

For our induction step, by assuming that $y_{n + 1} \leq y_{n}$, we then have

$$\begin{aligned} y_{n + 2} \leq y_{n + 1} &\iff \frac{(x_{n + 1} + y_{n + 1})}{2} \leq y_{n + 1} \\ &\iff x_{n + 1} + y_{n + 1} \leq 2y_{n + 1} \\ &\iff x_{n + 1} \leq y_{n + 1}. \end{aligned}$$

Again, the last inequality is true by applying the proven identity in (a). Since we've previously proven that all $y_{n}$ terms are positive, we can use 0 as its lower bound. This then allows us to apply MCT on $y_{n}$, assuring us of a limit this sequence. In finding an upper bound for $x_{n}$, notice that (this is why it was important to prove that $y_{n}$ is decreasing)

$$\forall n \in \mathbb{N}, \quad x_1 \leq x_2 \leq \dots \leq x_{n - 1} \leq x_{n} \leq y_{n} \leq y_{n - 1} \leq \dots \leq y_2 \leq y_1.$$

Therefore, we can use $y_1$ as an upper bound for all terms in the sequence $x_{n}$. Again, having satisfied the conditions of MCT for $x_{n}$, we know that $x_{n}$ converges as well. Now, to prove that both sequences converge to the same limit, notice that

$$\begin{aligned} \lim_{n \to \infty} y_{n} = \lim_{n \to \infty} \frac{x_{n} + y_{n}}{2} &\implies \lim_{n \to \infty} y_{n} = \frac{\lim_{n \to \infty} x_{n} + \lim_{n \to \infty} y_{n} }{2} \\ &\implies 2 \lim_{n \to \infty} y_{n} = \lim_{n \to \infty} x_{n} + \lim_{n \to \infty} y_{n} \\ &\implies \lim_{n \to \infty} y_{n} = \lim_{n \to \infty} x_{n}. \end{aligned}$$