Is this an acceptable proof that $x^2<y^2$ when $x$ and $y$ are positive reals with $x<y$

Question

The question is

Prove the following statement:

Suppose $x, y \in \mathbb{R}$ and that $x$ and $y$ are positive. If $x < y$ then $x^{2} < y^{2}$.

Proof:

Suppose there is a value $r$ between $x$ and $y$ so that
$$x < r < y \tag1$$ where $r \in \mathbb{R}$ and $ r > 0$. Therefore,
$$x<r \Rightarrow x^{2}<xr \tag2$$ and
$$r<y \Rightarrow yr < y^{2} \tag3$$
Moreover, if we multiply equation $(1)$ by $r$, we get
$$xr < r^{2} < yr \tag4$$
Combining equations $(2)$, $(3)$, and $(4)$, we get
$$x^{2} < xr < r^{2} < xy < y^{2} \tag5$$
Therefore, $x^{2} < y^{2}$. $\blacksquare$

If necessary, we can even define $r=$ $x+y\over{2}$

Is there anything wrong with this approach? I'm aware of the normal method of direct proof.

Answer 1

No need to introduce $r$; to wit:

$x = x; \tag 1$

$x < y; \tag 2$

thus

$x^2 < xy; \tag 3$

also,

$y = y; \tag 4$

from (2) and (4),

$xy < y^2; \tag 5$

combining (3) and (5):

$x^2 < xy < y^2. \tag 5$

$OE\Delta$.