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I am having problems solving this particular system of equations (SOE). The only thing I see that could be useful is the first and second equations are reciprocals of each other. I just don't know how I can use that information to help solve the SOE.

$$ \left\{ \begin{array}{c} \displaystyle\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}}=3 \\ \displaystyle\sqrt{\frac{y\:}{x\:}}+\sqrt{\frac{z\:}{y\:}}+\sqrt{\frac{x\:}{z\:}}=3 \\ \displaystyle\sqrt{xyz}=1 \end{array} \right. $$

Is there a way to solve this algebraically other than solving by inspection? Thank you for any help/assistance you can provide!

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  • $\begingroup$ Square bth sides of the last equation to get $x=\frac1{yz}$. Now eliminate $x$ from the first two equations, and see what you can do from there. $\endgroup$ – saulspatz Feb 6 at 6:55
  • $\begingroup$ A quick way of solving this problem is noticing that the first two lines are reciprocals of each other, in particular the unknowns. So you can make a substitution and rewrite the first two lines as $a+b+c=3$ and $a^{-1}+b^{-1}+c^{-1}=3$. The important observation here is the constant $3$ is left unchanged even though the unknowns are reciprocals. That would mean the unknowns must be $1$ because there is no other real number whose multiplicative inverse is itself. The third line also confirms the unknowns are all equal to $1$. $\endgroup$ – user314 Feb 6 at 11:20
  • $\begingroup$ Note that $-1$ is its own multiplicative inverse as well. However, I have shown that $-1$ doesn't work, though $\endgroup$ – Some Guy Feb 8 at 3:21
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Note that for $x,y,z >0$ ,we have $$\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{z}}+\sqrt{\dfrac{z}{x}} \geq 3$$ By the AM-GM inequality, and it is given that it is equal to it's minimum value, and that happens only when all the terms are equal, that is, $$\sqrt{\dfrac{x}{y}}=\sqrt{\dfrac{y}{z}}=\sqrt{\dfrac{z}{x}}$$ We can see that now $x=y=z=1$ is the only solution possible in $\mathbb{R}$ which is satisfied by other equations as well.

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You can set $\sqrt{x/y}=a$, $\sqrt{y/z}=b$, $\sqrt{z/x}=c$, so the first two equations become $$ a+b+c=3,\qquad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3 $$ and also $abc=1$. Hence the second equation yields $ab+bc+ca=3$.

Therefore the three numbers are the roots of the polynomial $t^3-3t^2+3t-1$. Now you can finish, can't you?

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Unless you are looking for a generalization with RHS other than $3$ and $1$, this system of equations is solvable by inspection where $$x=y=z=1$$

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