Find $a,b,c \in \mathbb R$ such that: $$\left\{ \begin{align} a+b+c &=4 \\ \left( a+b \right) \left( b+c \right) \left( c+a \right) &=18 \\ \frac 1 a+\frac 1b+\frac 1c &=\frac 52 \end{align} \right.$$
How do I solve for $a,b$ and $c$ manually (without using a computer) ? Thanks !
All solutions are $(1,1,2)$ and permutations
$$ \begin{align} a+b+c &=4 \\ \left( a+b \right) \left( b+c \right) \left( c+a \right) &=18 \\ \frac 1 a+\frac 1b+\frac 1c &=\frac 52 \end{align} $$
It is useful to notice that all expressions here are expressible using symmetric polynomials and, in turn, they are expressible using the elementary symmetric polynomials. $S_1=a+b+c$, $S_2=ab+ac+bc$, $S_3=abc$
First equation is simply $S_1=4$.
From the second equation we get $(a+b)(b+c)(c+a)=2abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2=(ab+ac+bc)(a+b+c)-abc=S_1S_2-S_3=18$.
The third one gives $\frac 1a+\frac 1b+\frac 1c =\frac 52$ $\Leftrightarrow$ $\frac{ab+ac+bc}{abc}=\frac{S_2}{S_3}=\frac52$.
So we have $$\begin{align} S_1&=4\\ S_1S_2-S_3&=18\\ S_2&=\frac52S_3 \end{align}$$ which has the solution $S_1=4$, $S_2=5$, $S_3=2$.
Now using Vieta's formulas we get that $a$, $b$, $c$ are solutions of the cubic equation $$x^3-4x^2+5x-2=0.$$
Now we can apply rational roots test, which reveals that $x=1$ is a solution and then we get $x^3-4x^2+5x-2=(x-1)(x^2-3x+2)=(x-1)^2(x-2)$. (Or we could employ one of the methods used for solving cubic equations.)
We can check the solution or WolframAlpha or simply verify that $a=b=1$, $c=2$ fulfill the original system. (Of course, the permutations are solutions of the original equations, too.)