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I have the problem Find all triples $(a,b,c)$ of real numbers such that the following system holds:

$$a+b+c=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \\a^2+b^2+c^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}$$

I tried

Solve[{a + b + c == 1/a + 1/b + 1/c,    a^2 + b^2 + c^2 == 1/a^2 + 1/b^2 + 1/c^2}, {a, b, c}, Reals]

I don't get the result.

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  • $\begingroup$ Mathematica v12.2 evaluates your Solve command! $\endgroup$ Commented Dec 11, 2023 at 7:59
  • $\begingroup$ Solve[{a + b + c == 1/a + 1/b + 1/c, a^2 + b^2 + c^2 == 1/a^2 + 1/b^2 + 1/c^2}, Reals] $\endgroup$
    – cvgmt
    Commented Dec 11, 2023 at 8:05
  • $\begingroup$ When you say you don't "get" the result, do you mean Mathematica doesn't produce one, or that you don't understand it? My $Version=="13.2.0 for Mac OS X x86 (64-bit) (November 18, 2022)" produces a lengthy result with lots of solutions just running your command as presented. But the important thing is that the solutions provided are not tuples of explicit numbers, but they are parametric. They have to be parametric, as you have 3 variables but only 2 polynomial equations. $\endgroup$
    – evanb
    Commented Dec 11, 2023 at 9:58

1 Answer 1

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Here an easy way to visualize the solutions:

abc = {a, b, c} /. 
FindInstance[{ a + b + c == 1/a + 1/b + 1/c &&a^2 + b^2 + c^2 == 1/a^2 + 1/b^2 + 1/c^2
,-2 < a < 2, -2 <b < 2, -2 < c < 2}, {a, b, c}, Reals, 1000] // N;
Show[Graphics3D[{Point[abc]}], AxesLabel -> {a, b, c}, Axes -> True]

enter image description here

As mentioned in my comment Solve evaluates too:

abc = {a, b, c} /. 
Solve[{a + b + c == 1/a + 1/b + 1/c,a^2 + b^2 + c^2 == 1/a^2 + 1/b^2 + 1/c^2}, {a, b, c}, Reals];

Show[{ParametricPlot3D[Evaluate[abc[[1 ;; 4]]], {b, -2, 2}], 
ParametricPlot3D[Evaluate[abc[[5 ;; 10]]], {a, -2, 2}], 
Graphics3D[{PointSize[Medium], Point[abc[[11 ;; 18]]]}]}, 
BoxRatios -> {1, 1, 1}, AxesLabel -> {a, b, c} ,PlotRange -> {{-2, 2}, {-2, 2}, {-2, 2} }]  

enter image description here

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