Solve $a=\frac{b+c}{1+b^2c^2},b=\frac{a+c}{1+a^2c^2}, c=\frac{a+b}{1+a^2b^2}$

Question

Let $a,b,c> 0$ such that $$a=\frac{b+c}{1+b^2c^2},b=\frac{a+c}{1+a^2c^2}, c=\frac{a+b}{1+a^2b^2}$$

Prove that the only solution to this system of equations is $$a=b=c=1$$

I am getting this answer using numerical simulations, but haven't been able to prove this algebraically.

EDIT: I am specifically looking for an argument that utilizes inequalities, and is an equality only for $a=b=c=1$.

Answer 1

Let $s = a+b+c$ and $p = abc$. Then, we have $$\begin{align}a = \dfrac{b+c}{1+b^2c^2} &= \dfrac{s-a}{1+\tfrac{p^2}{a^2}} \\ a\left(1+\dfrac{p^2}{a^2}\right) &= s-a \\ a^2+p^2 &= sa-a^2 \\ 2a^2-sa+p^2 &= 0 \\ a &= \dfrac{s + \epsilon_a\sqrt{s^2-8p^2}}{4},\end{align}$$ where $\epsilon_a = +1$ or $\epsilon_a = -1$, and similarly for $b$ and $c$. Hence, $$\begin{align}s = a+b+c &= \dfrac{3s + (\epsilon_a+\epsilon_b+\epsilon_c)\sqrt{s^2-8p^2}}{4} \\ s &= (\epsilon_a+\epsilon_b+\epsilon_c)\sqrt{s^2-8p^2}\end{align}$$

Since $a,b,c, > 0$, we have $s > 0$, and so we can only have $\epsilon_a+\epsilon_b+\epsilon_c = 1$ or $3$.

If $\epsilon_a+\epsilon_b+\epsilon_c = 1$, then we get $s = \sqrt{s^2-8p^2}$, which yields $p = 0$, which isn't possible since $a,b,c > 0$.

If $\epsilon_a+\epsilon_b+\epsilon_c = 3$, then we have $\epsilon_a = \epsilon_b = \epsilon_c = +1$, i.e. $a = b = c$. Substituting yields $a = \dfrac{b+c}{1+b^2c^2} = \dfrac{2a}{1+a^4}$, and solving yields $a = b = c = 1$ (since $a,b,c > 0$).

Therefore, $a = b = c = 1$ is the only positive solution.

Answer 2

Another solution by substitution :

Substitute $x=bc$, $y=ca$, $z=ab$. Then we get the following.

$$\frac{1}{y}+\frac{1}{z}=\frac{1}{x}+x$$

$$\frac{1}{z}+\frac{1}{x}=\frac{1}{y}+y$$

$$\frac{1}{x}+\frac{1}{y}=\frac{1}{z}+z$$

This implies the following.

$$\frac{2}{x}+x=\frac{2}{y}+y = \frac{2}{z}+z $$

By the property of second-order polynomial equation, at least two of $x,y,z$ should be same. WLOG $x=y$, and we get $xz=yz=1$.

Finally, we get $x=\frac{1}{x}$, and $x=y=z=1$. This implies $a=b=c=1$. It is easy to check that this is indeed a unique solution.

Answer 3

From the system of equations, we have \begin{align*} a^2 + a^2b^2c^2 &= a(b + c), \tag{1}\\ b^2 + a^2b^2c^2 &= b(c + a), \tag{2}\\ c^2 + a^2b^2c^2 &= c(a + b). \tag{3} \end{align*}

$(1) - (2)$ yields $$(a - b)(a + b - c) = 0. \tag{4}$$

$(2) - (3)$ yields $$(b - c)(b + c - a) = 0. \tag{5}$$

We claim that $a = b$. Indeed, if $a + b - c = 0$, we have $c = a + b$, which, when plugged into (5), gives $-2ab = 0$ which is impossible.

From $a = b$ and (5), we have $(b-c)c = 0$ which results in $b = c$.

From (1) and $a = b = c$, we have $a^6 = a^2$ which results in $a = 1$.

We are done.