Tintarn wrote: Call a triple $(a,b,c)$ of positive numbers mysterious if \[\sqrt{a^2+\frac{1}{a^2c^2}+2ab}+\sqrt{b^2+\frac{1}{b^2a^2}+2bc}+\sqrt{c^2+\frac{1}{c^2b^2}+2ca}=2(a+b+c).\]Prove that if the triple $(a,b,c)$ is mysterious, then so is the triple $(c,b,a)$. Proposed by A. Kuznetsov, K. Sukhov If $abc<1$ we have \begin{align*} \sqrt{a^2+\frac{1}{a^2c^2}+2ab}+\sqrt{b^2+\frac{1}{b^2a^2}+2bc}+\sqrt{c^2+\frac{1}{c^2b^2}+2ca}\\ >\sqrt{a^2+b^2+2ab}+\sqrt{b^2+c^2+2bc}+\sqrt{c^2+a^2+2ca}\\ =(a+b)+(b+c)+(c+a)=2(a+b+c). \end{align*}Analogously $\sqrt{a^2+\frac{1}{a^2c^2}+2ab}+\sqrt{b^2+\frac{1}{b^2a^2}+2bc}+\sqrt{c^2+\frac{1}{c^2b^2}+2ca}=2(a+b+c)$ for $abc=1$ and $\sqrt{a^2+\frac{1}{a^2c^2}+2ab}+\sqrt{b^2+\frac{1}{b^2a^2}+2bc}+\sqrt{c^2+\frac{1}{c^2b^2}+2ca}<2(a+b+c)$ for $abc>1$. So $(a,b,c)$ is mysterious iff $abc=1$. In particular if $(a,b,c)$ is mysterious, so is $(c,b,a)$.

Tintarn wrote: Call a triple $(a,b,c)$ of positive numbers mysterious if \[\sqrt{a^2+\frac{1}{a^2c^2}+2ab}+\sqrt{b^2+\frac{1}{b^2a^2}+2bc}+\sqrt{c^2+\frac{1}{c^2b^2}+2ca}=2(a+b+c).\]Prove that if the triple $(a,b,c)$ is mysterious, then so is the triple $(c,b,a)$. Proposed by A. Kuznetsov, K. Sukhov Interesting.

I like how the actual underlying structure of this problem is disguised. It took me a good while to realize what was actually going on. The entirety of the proof is the following result. Claim : A triple of positive reals $(a,b,c)$ is mysterious if and only if $abc=1$. Proof : We let $abc=k \in \mathbb{R}_{>0}$. Then, \[S=\sqrt{a^2+\frac{1}{a^2c^2}+2ab}+\sqrt{b^2+\frac{1}{b^2a^2}+2bc}+\sqrt{c^2+\frac{1}{c^2b^2}+2ca}=\sqrt{a^2+\frac{b^2}{k^2}+2ab}+\sqrt{b^2+\frac{c^2}{k^2}+2bc}+\sqrt{c^2+\frac{a^2}{k^2}+2ca}\]We proceed by considering two cases. Case 1 : $0<k<1$ ($0<k^2<1$). Here, $\frac{a^2}{k^2}>a^2$ and likewise. Thus, \begin{align*} S &= \sqrt{a^2+\frac{b^2}{k^2}+2ab}+\sqrt{b^2+\frac{c^2}{k^2}+2bc}+\sqrt{c^2+\frac{a^2}{k^2}+2ca}\\ & > \sqrt{a^2+b^2+2ab}+\sqrt{b^2+c^2+2bc}+\sqrt{c^2+a^2+2ac}\\ &= (a+b) + (b+c) + (c+a)\\ &= 2(a+b+c) \end{align*}Thus, $S>2(a+b+c)$ which contradicts the desired equality. Case 2 : $k>1$. Then, $0 < \frac{a^2}{k^2}<a^2$ and likewise. Thus, \begin{align*} S & =\sqrt{a^2+\frac{b^2}{k^2}+2ab}+\sqrt{b^2+\frac{c^2}{k^2}+2bc}+\sqrt{c^2+\frac{a^2}{k^2}+2ca}\\ & < \sqrt{a^2+b^2+2ab} + \sqrt{b^2+c^2+2bc} + \sqrt{c^2+a^2 +2ac}\\ &= (a+b) + (b+c) + (c+a)\\ &= 2(a+b+c) \end{align*}So, $S<2(a+b+c)$ which again contradicts the given equation. Thus, if $abc \ne 1$, then the triple $(a,b,c)$ cannot be mysterious. The other direction is trivial, simply substitute $k=1$ to the above equation and simplify. This proves the claim. Now, the problem statement is clear since due to the associativity of real multiplication, $1=abc=cba$ so if $(a,b,c)$ is mysterious so is $(c,b,a)$.