Prove $\forall\ x,\ y \in \left]-1,1\right[: \frac{x+y}{1+xy} \in \left]-1,1\right[$ using only elementary algebraic transformations

Question

I have to prove the following statement:

$$\forall\ x,\ y \in \left]-1,1\right[: \frac{x+y}{1+xy} \in \left]-1,1\right[$$

A way to prove it is via analysing the behaviour of $\ f(x,y)=\frac{x+y}{1+xy}\ $ in $\left]-1,1\right[.$

Can anyone support me with any hints how to prove it using only elementary algebraic transformations.

Answer 1

$$\frac{x+y}{1+xy}+1=\frac{1+xy+x+y}{1+xy}=\frac{(1+x)(1+y)}{1+xy}>0$$ and $$\frac{x+y}{1+xy}-1=-\frac{1+xy-x-y}{1+xy}=-\frac{(1-x)(1-y)}{1+xy}<0,$$ which says $$-1<\frac{x+y}{1+xy}<1.$$

Answer 2

So we need to prove that $$|x+y|<|1+xy|$$ or $$(x+y)^2< (1+xy)^2$$ or $$ x^2+y^2< 1+x^2y^2$$ or $$ \underbrace{(x^2-1)}_{< 0}\cdot \underbrace{(y^2-1)}_{< 0}> 0$$ which is true by assumption that $|x|< 1$ and $|y|<1$

Answer 3

Note that $xy \in \left]-1,1\right[\,$, so $1+xy \gt 0\,$, then prove that $-1-xy \lt x+y \lt 1+xy\,$.

Hint for the leftmost inequality: $\;(1+x)(1+y) \gt 0\,$.

Answer 4

Hint:

$$xy+x+y+1=(x+1)(y+1)$$ $$xy-x-y+1=(x-1)(y-1)$$

Answer 5

ok, Here's a more concrete version of the answer I posted earlier.

Transform $x$ and $y$ thus: $x\mapsto \dfrac{1-x}{1+x}= u$ and $y\mapsto \dfrac{1-y}{1+y} = v.$

Show that what you get is two positive numbers $u,v.$

Multiply them. What you get from multiplying two positive numbers is a positive number $uv.$

Transform $uv$ in the same way: $uv\mapsto \dfrac{1-uv}{1+uv}= p.$

Show that that gets you a number between $-1$ and $1.$

Finally, show that $p = \dfrac{x+y}{1+xy}.$

Below you see the slightly less concrete version that I posted earlier.

$$ \S \qquad \S \qquad \S \qquad \S \qquad \S $$

The mapping $x\mapsto \dfrac{1-x}{1+x}$ transforms this operation on the interval $(-1,1)$ to ordinary multiplication on the interval $(0,\infty).$

Let $x\diamond y = \dfrac{x+y}{1+xy}.$

Let $x^\dagger= \dfrac{1-x}{1+x}.$

This operation $x\mapsto x^\dagger$ takes $(-1,1)$ to $(0,\infty)$ thus: $\begin{cases} \phantom{-}1\mapsto 0 \\ \phantom{-}0\mapsto 1 \\ -1\mapsto\infty \end{cases}$

And $(x^\dagger)^\dagger = x.$

\begin{align} (x\diamond y)^\dagger & = x^\dagger y^\dagger \\[10pt] (xy)^\dagger & = x^\dagger\diamond y^\dagger \end{align}

So ordinary multiplication on $(0,\infty)$ and this $\diamond$-operation are conjugate to each other.

Since $(0,\infty)$ is closed under multiplication, $(-1,1)$ is therefore closed under this $\diamond$-operation.

Answer 6

I don't know if you classify the following as an elementary transformation, but I'll try anyway. Let $x=\tanh\alpha$ and $y=\tanh\beta$ which are automatically in $\left(-1,1\right)$. Then

$$\frac{x+y}{1+xy}=\tanh\left(\alpha+\beta\right)$$

is also in $\left(-1,1\right)$. The edge cases are easy.