If $x+y+z=xyz$, prove $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$

Question

If $x+y+z=xyz$, prove $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$ given that $x^2~,~y^2~,~z^2\ne1$

I came across this question in an ancient ($19$th century) Trigonometry book, and this is the method they use to prove the result (please note: I understand this method fully): Let $x=\tan A$ ,$~y=\tan B$ and $z=\tan C$ which is acceptable without any loss of generality.

This means that we are saying that $$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$ Consider $\tan (A+B+C)$: $$\tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1- \tan A \tan B- \tan C \tan A - \tan B \tan C}$$ So if $~\tan A+\tan B+\tan C=\tan A\tan B\tan C~~$ then $~~\tan (A+B+C)=0$. Hence, let $~A+B+C=\pi$.

Now consider $\tan (2A+2B+2C)$: $$\tan (2A+2B+2C)=\frac{\tan 2A+\tan 2B+\tan 2C-\tan 2A\tan 2B\tan 2C}{1- \tan 2A \tan 2B- \tan 2C \tan 2A - \tan 2B \tan 2C}=0$$ $$\implies \tan 2A+\tan 2B+\tan 2C=\tan 2A\tan 2B\tan 2C$$ $$\implies \frac{2\tan A}{1-\tan^2 A}+\frac{2\tan B}{1-\tan^2 B}+\frac{2\tan C}{1-\tan^2 C}=\frac{2\tan A}{1-\tan^2 A}\times\frac{2\tan B}{1-\tan^2 B}\times\frac{2\tan C}{1-\tan^2 C}$$ $$\therefore\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$$ as required.

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My question is, is there any other way of proving this without this rather heavy use of trigonometry? I also would prefer not to work through heaps of algabraic manipulation and expansion to obtain the required result, although if that's necessary I will put up with it ;)

Thank you for your help.

Answer 1

Define a new operation $p \otimes q = \frac{{p + q}}{{1 - pq}}$.

And it's easy to find that it's associative.

$p \otimes q \otimes r = \frac{{p + q + r - pqr}}{{1 - pq - pr - qr}}$.

It means $p + q + r = pqr \Leftrightarrow p \otimes q \otimes r = 0$.

And then $\begin{array}{l} \frac{{2x}}{{1 - {x^2}}} \times \frac{{2y}}{{1 - {y^2}}} \times \frac{{2z}}{{1 - {z^2}}} - (\frac{{2x}}{{1 - {x^2}}} + \frac{{2y}}{{1 - {y^2}}} + \frac{{2z}}{{1 - {z^2}}}) = 0\\ \Leftrightarrow \frac{{2x}}{{1 - {x^2}}} \otimes \frac{{2y}}{{1 - {y^2}}} \otimes \frac{{2z}}{{1 - {z^2}}} = 0\\ \Leftrightarrow (x \otimes x) \otimes (y \otimes y) \otimes (z \otimes z) = 0\\ \Leftrightarrow (x \otimes y \otimes z) \otimes (x \otimes y \otimes z) = 0\\ \Leftrightarrow 0 \otimes 0 = 0 \end{array}$

So the problem is solved.

Answer 2

$$2x(1-y^2)(1-z^2)=2x-2xy^2-2xz^2+2xy^2z^2$$

$$2xy^2z^2=2yz(x+y+z)=2xyz + 2y^2z+2yz^2$$

$$\implies 2x(1-y^2)(1-z^2)=2x-2xy^2-2xz^2+2xyz+2y^2z+2yz^2$$ $$\implies 2y(1-z^2)(1-x^2)=2y-2yz^2-2yx^2+2xyz+2z^2x+2zx^2$$ $$\implies 2z(1-x^2)(1-y^2)=2z-2zx^2-2zy^2+2xyz+2x^2y+2xy^2$$

Summing up gives $2(x+y+z+3xyz)=8xyz$, which reduces to the equality.

I wish to see a simpler proof as well.

Answer 3

Hint:

$$x+y+z=xyz\iff x=\dfrac{y+z}{1-yz}$$

If $\dfrac{2x}{1-x^2}=a$ etc. assuming $x^2,y^2,z^2\ne1$, we need $$a+b+c=abc\iff\dfrac1{bc}+\dfrac1{ca}+\dfrac1{ab}=1$$

$$\implies\dfrac1a\left(\dfrac1b+\dfrac1c\right)$$

$$=\dfrac{(1-x^2)}{2x}\left(\dfrac{1-y^2}{2y}+\dfrac{1-z^2}{2z}\right)$$

$$=\dfrac{1-\left(\dfrac{y+z}{1-yz}\right)^2}{2\left(\dfrac{y+z}{1-yz}\right)}\cdot\dfrac{(1-yz)(y+z)}{2yz}$$

$$=\dfrac{(1-yz)(y+z)((1-yz)^2-(y+z)^2)}{4(y+z)(1-yz)yz}$$

$$=\dfrac{4yz-(1-y^2)(1-z^2)}{4yz}$$ $$=1-\left(\dfrac{1-y^2}{2y}\right)\left(\dfrac{1-z^2}{2z}\right)$$

$$=1-\dfrac1{bc}$$

So, we are done!

Answer 4

Multiplying the eq. to prove by $\frac{(1-x^2)(1-y^2)(1-z^2)}{2}$ you obtain :

$$\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2} \iff$$ $$x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)= 4xyz$$

but expanding the left side under the hypothesis $x+y+z=xyz\ $, you get : $$x y^2 z^2 - x y^2 - x z^2 + x +x^2 y z^2 - x^2 y - y z^2 + y+ x^2 y^2 z - x^2 z - y^2 z + z =$$ $$x+y+z+xyz(yz+xy+xz)- x y^2 - x z^2- x^2 y - y z^2 - x^2 z - y^2 z=$$ $$(x+y+z)(yz+xy+xz+1)- x y^2 - x z^2- x^2 y - y z^2 - x^2 z - y^2 z=$$ $$xyz+xyz+xyz+x+y+z=3xyz+x+y+z = 4xyz$$

Answer 5

Consider the polynomial ring $\mathbb{Z}[X, Y, Z]$ with integer coefficients in three indeterminates. For simplicity denote $s\colon=X+Y+Z$, $r\colon=XY+YZ+ZX$ and $p\colon=XYZ$, the three fundamental symmetric polynomials in $X, Y, Z$.

A careful expansion reveals that: $$\begin{align*} &X\left(1-Y^2\right)\left(1-Z^2\right)+Y\left(1-Z^2\right)\left(1-X^2\right)+Z\left(1-X^2\right)\left(1-Y^2\right)=\\ &s+XY^2Z^2+YZ^2X^2+ZX^2Y^2-XY^2-YX^2-YZ^2-ZY^2-ZX^2-XZ^2\\ &=s+rp-XY(X+Y)-YZ(Y+Z)-ZX(X+Z)\\ &=s+rp-XY(s-Z)-YZ(s-X)-ZX(s-Y)\\ &=s+rp-sr+3p\\ &=s+r(p-s)+3p \end{align*}$$ Introducing the principal ideal $I\colon=(p-s) \subseteq \mathbb{Z}[X, Y, Z]$, it is clear that $s+r(p-s)+3p \equiv 4p\ (\mathrm{mod}\ I)$. If we introduce the factor ring $A\colon=\mathbb{Z}[X, Y, Z]/I$ and denote the images of the indeterminates $X, Y, Z$ with $x, y, z$ we then obtain: $$x\left(1-y^2\right)\left(1-z^2\right)+y\left(1-z^2\right)\left(1-x^2\right)+z\left(1-x^2\right)\left(1-y^2\right)=4xyz,$$ which after multiplying by $2$ takes the form of the equality between the numerators in your fraction (obtained after taking $\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)$ as a common denominator).

Any ring $B$ containing three pairwise permutable elements $u, v, w$ such that $u+v+w=uvw$ constitutes itself as the codomain of a unique ring morphism defined on $A$ and mapping $x \mapsto u$, $y \mapsto v$ and $z \mapsto w$.

The moral is that for any such conditional algebraic identity there must be an underlying universal, polynomial identity or congruence.

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Appendix. The answer of @sraung Jo contains an indeed remarkable idea, which however enjoys only superficial and not entirely rigorous treatment. As compared to the previous paragraph, I believe you will find this alternative version to be quite finicky with the details, although one can argue it affords a considerable insight of an algebro-geometric nature. Otherwise put, when judged by mere concision, the universal polynomial congruences presented above still constitute the simplest and most direct answer.

Consider an arbitrary commutative field $K$ subject to the sole condition that $-1_K$ is not a square in $K$. I will make here the explicit mention that I do not follow the English custom of indicating the distinction between commutative and non-commutative fields by referring to the latter as "division rings", for I find this custom quite foolish from the taxonomical point of view (about as foolish as referring to "wide-beaked birds" as mere birds, and to "not necessarily wide-beaked birds" as "flying animals").

Consider a fixed object that does not belong to set $K$ -- the precise formal justification that guarantees the existence of such an object will vary depending on the axiomatic system&formal language one uses to express mathematics, but suffice to say that in most standard systems this is possible -- which we shall denote as $\infty_K$ and consider the extended set $\tilde{K}\colon=K \cup \{\infty_K\}$, typically known as the projective completion of $K$.

On this completion we shall define the following binary operation: $$\begin{align} *\colon \tilde{K} \times \tilde{K} &\to \tilde{K}\\ x*y&=\begin{cases} \frac{x+y}{1_K-xy}, &x, y \in K \wedge xy \neq 1_K\\ \infty_K, &x, y \in K \wedge xy=1_K\\ -\frac{1}{y}, &x=\infty_K \wedge y \in K^{\times}\\ \infty_K, &x=\infty_K \wedge y=0_K\\ -\frac{1}{x}, &x \in K^{\times} \wedge y=\infty_K\\ \infty_K, &x=0_K \wedge y=\infty_K\\ 0_K, &x=y=\infty_K. \end{cases} \end{align}$$

It becomes a matter of interest to study the elementary properties of the magma $(\tilde{K}, *)$ (by magma I mean a pair $(A, \cdot)$ consisting of a set $A$ together with an abstract binary operation $\cdot \colon A \times A \to A$, subject to no axioms whatsoever). It turns out that:

• $*$ is an associative operation, claim whose justification we omit here, as it is not at all difficult but merely tedious (there are several cases that have to be handled separately). This is where the condition that $-1_K$ is not a square comes into play
• $*$ admits $0_K$ as a neutral element, as can be easily ascertained from the definition
• every element $x \in \tilde{K}$ has a symmetric with respect to $*$: for $x \in K$ we have $x*(-x)=(-x)*x=0_K$ -- which means that $-x$ is the symmetric of $x$ -- and similarly $\infty_K *\infty_K=0_K$, meaning that $\infty_K$ is its own inverse
• $*$ is commutative, fact which follows right away from the definition.

The above observations signify that $(\tilde{K}, *)$ is actually an abelian group. It is also easy to see that $\infty_K$ is the only element of order $2$ in $\tilde{K}$, respectively that $\pm1_K$ are the only elements of order $4$ and they generate the same cyclic subgroup $\{1_K, \infty_K, -1_K,0_K\} \leqslant \tilde{K}$ which is thus established as the unique cyclic subgroup of order $4$ of $\tilde{K}$.

Let us make a brief remark on what the structure of this group is in the particular case $K=\mathbb{R}$. We consider the map: $$\begin{align} \tau \colon \mathbb{R} &\to \tilde{\mathbb{R}}\\ \tau(x)&=\begin{cases} \mathrm{tg}\left(\pi x\right), &x \notin \mathbb{Z}+\frac{1}{2}\\ \infty_{\mathbb{R}}, &x \in \mathbb{Z}+\frac{1}{2} \end{cases} \end{align}$$ and argue that it is a surjective morphism between the groups $(\mathbb{R}, +)$ and $(\tilde{\mathbb{R}}, *)$. As the kernel is given by $\mathrm{Ker}\tau=\mathbb{Z}$, we infer from the fundamental homomorphism theorem that $\tilde{\mathbb{R}} \approx \mathbb{R}/\mathbb{Z} \hspace{3pt} (\mathbf{Gr})$, in other words the group $(\tilde{\mathbb{R}}, *)$ is isomorphic to the circle group $\mathbb{U}\colon=\{z \in \mathbb{C}|\ |z|=1\}$.

As to the conditional identity you are interested in, while it is true that this general group structure does have a relation to it, the equivalence $p+q+r=pqr \Leftrightarrow p*q*r=0_K$ glibly asserted in the answer referenced above does not in general hold and -- furthermore -- analysing the explicit connection between the conditions $x+y+z=xyz$ and $x*y*z=0_K$ for three elements $x, y, z \in K$ requires dealing with several singular cases described by algebraic equations over $K$, therefore not being the most recommendable means to prove your desired identity. As far as the group structure $(\tilde{K}, *)$ itself is concerned, it is indeed worthy of all consideration.

Answer 6

Let $z=0$.

Thus, $x+y=0$ and $$\sum_{cyc}\frac{2x}{1-x^2}=\frac{2x}{1-x^2}-\frac{2x}{1-x^2}=0=\prod_{cyc}\frac{2x}{1-x^2}.$$ Now, let $xyz\neq0.$

Thus, we need to prove that: $$\sum_{cyc}\frac{1}{\frac{2x}{1-x^2}\cdot\frac{2y}{1-y^2}}=1.$$ Indeed, $$\sum_{cyc}\frac{1}{\frac{2x}{1-x^2}\cdot\frac{2y}{1-y^2}}=\sum_{cyc}\frac{(1-x^2)(1-y^2)}{4xy}=\sum_{cyc}\frac{(\frac{xyz}{x+y+z}-x^2)(\frac{xyz}{x+y+z}-y^2)}{4xy}=$$ $$=\sum_{cyc}\frac{(yz-x(x+y+z))(xz-y(x+y+z))}{4(x+y+z)^2}=$$ $$=\sum_{cyc}\frac{x^2yz-(x+y+z)(x^2y+x^2z)+xy(x+y+z)^2}{4(x+y+z)^2}=$$ $$=\sum_{cyc}\frac{\frac{1}{3}xyz-x^2y-x^2z+xy(x+y+z)}{4(x+y+z)}=\sum_{cyc}\frac{\frac{1}{3}xyz+xyz}{4xyz}=1.$$