We claim that x=y=z
Suppose to the contrary, we have say x>z, x+y>y+z \Rightarrow z =(x+y)^3 >(y+z)^3>x, a contradiction.
Hence the two solutions given above are the only solutions
find all real solutions of the system
(x+y)^{3}=z\, ,(y+z)^{3}=x\, ,(z+x)^{3}=y
We claim that x=y=z
Suppose to the contrary, we have say x>z, x+y>y+z \Rightarrow z =(x+y)^3 >(y+z)^3>x, a contradiction.
Hence the two solutions given above are the only solutions
so sir inderiectly we can say the cyclic +symetric function obtain optimum at x=y=z
Not necessary. The simplest counter-example I could offer would be:
Find all pairs (x,y) satisfying
\begin{matrix} x^2=y\\ y^2=x \end{matrix}
Assuming x = y would lead us away from two other solutions which are ...?
the point here was that you cannot exhaust all possible solutions with the assumption that the variables are equal.
Here, we had to prove that no other solutions exist other than when the variables are all equal.
u missed this solution:
\left\{\frac{-1}{2\sqrt{2}},\frac{-1}{2\sqrt{2}},\frac{-1}{2\sqrt{2}} \right\}