Solving a symmetric equation involving three variables a,b and c

Question

Solve : $$\frac{x+a^2}{a+b}+\frac{x+b^2}{b+c}+\frac{x+c^2}{c+a}=2(a+b+c)$$ I am trying to find a simple technique to solve this equation as there is a pattern in the equation, but I could not do. Any help will be highly appreciated.

My attempt: I tried the long method by separating all $x$ terms on the Left Hand Side and the other terms on the Right Hand Side, but even that is becoming too big. Please suggest an appropriate method.

Answer 1

Expanding and factoring immediately gives $$ x = ab+ac+ bc. $$

Answer 2

$$\begin{align}\frac{x+a^2}{a+b}+\frac{x+b^2}{b+c}+\frac{x+c^2}{c+a}&=2(a+b+c)\\ \frac{x}{a+b}+\frac{a^2}{a+b}+\frac{x}{b+c}+\frac{b^2}{b+c}+\frac{x}{c+a}+\frac{c^2}{c+a}&=2(a+b+c)\\ x(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})&=\frac{2a(a+b)-a^2}{a+b}+\frac{2b(b+c)-b^2}{b+c}+\frac{2c(c+a)-c^2}{c+a}\\ x\left(\frac{a^2+b^2+c^2+3ab+3ac+3b}{(a+b)(b+c)(c+a)}\right)&=\frac{2a(a+b)-a^2}{a+b}+\frac{2b(b+c)-b^2}{b+c}+\frac{2c(c+a)-c^2}{c+a} \end{align} $$ I think this is nice enough to solve.

Answer 3

Let $\sum_{cyc}$ be cyclic sums over the 3 variables $a,b,c$. The equation at hand can be rewritten as

$$\sum_{cyc} \frac{x+a^2}{a+b} = 2\sum_{cyc}a$$

One approach to simplify this expression is convert it to cyclic sums that involve only one variable. Notice

$$\sum_{cyc}\frac{x+b^2}{a+b} = \sum_{cyc}\frac{(x+a^2)+(b^2-a^2)}{a+b} =\sum_{cyc}\left[\frac{x+a^2}{a+b} + (b-a)\right] = \sum_{cyc} \frac{x+a^2}{a+b}$$ We can rewrite the equation as

$$\sum_{cyc} \frac{2x + a^2 + b^2}{a + b} = 4\sum_{cyc} a$$

Let $\lambda = \sum_{cyc} a$ and $\mu = 2x + \sum_{cyc} a^2 - \lambda^2$, above equation becomes

$$\begin{align} & \sum_{cyc} \frac{\mu + \lambda^2 - c^2}{\lambda - c} = \sum_{cyc} \left[\frac{\mu}{\lambda - c} + (\lambda + c)\right] = 4\lambda\\ \implies & \mu \sum_{cyc}\frac{1}{\lambda - c} = 4\lambda - (3\lambda + \sum_{cyc} c ) = 0\\ \implies & 2x + \sum_{cyc} a^2 - \lambda^2 = \mu = 0\\ \implies & x = \frac12\left[\left(\sum_{cyc} a\right)^2 - \sum_{cyc} a^2 \right] = ab+bc+ca \end{align} $$

Answer 4

We have: $$\frac{x+a^2}{a+b}+\frac{x+b^2}{b+c}+\frac{x+c^2}{c+a} = 2(a+b+c) = (c+a) + (a+b) + (b+c)$$ $$\frac{x+a^2}{a+b}+\frac{x+b^2}{b+c}+\frac{x+c^2}{c+a} = \frac{(a+b)(c+a)}{a+b}+\frac{(a+b)(b+c)}{b+c}+\frac{(b+c)(c+a)}{c+a}$$ $$= \frac{a^2+(ab+bc+ca)}{a+b}+\frac{b^2+(ab+bc+ca)}{b+c}+\frac{c^2+(ab+bc+ca)}{c+a}$$ Now, we can see that $x=ab+bc+ca$ is one solution.

Let $x'$ be another solution such that $x'=x+d$. Then, substituting it in the equation, we have: $$ \frac{d}{a+b}+\frac{d}{b+c}+\frac{d}{c+a} = 0 $$ which implies: $$ \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} = 0 $$ However, the above need not be true for all $(a,b,c)$

Hence, we can conclude that the only solution is $x=ab+bc+ca$ unless $a^2+b^2+c^2+3ab+3bc+3ca = 0$ where any $x$ would work.

Note that you can find values $c$ for constants $a$ and $b$ to satisfy the above special criteria using quadratic equations. You can try it out!