$x$ and $y$ are real numbers (edited: thanks to @mcd's answer, now the "positive" requirement is removed) constrained by $ax+by=c$ where $a,b,c$ are strictly positive number.Given a strictly increasing, positive and continuous function $f(\cdot)$, if there exists solution(s) to $a\cdot f(x)+b\cdot f(y)=z$ where $z$ is a strictly positive number, prove that one of these solutions features that $x\geq y$.
A more direct way to see this is to use the graphs. The solutions to the equations are intersection points between lines $ax+by=c$ and $a\cdot f(x)+b\cdot f(y)=z$ (please see the following graphs). There are always intersection points falling in the bottom right half of the figure (that is, $x\geq y$).
I don't know how to prove this and I don't even know whether the claim above is right. But it seems intuitive when $a=b$ because in this case, if $x<y$ we can just exchange the values of $x$ and $y$ and we have $x\geq y$ and $x,y$ is a solution to the equations (see fig 1). But generally, when $a\neq b$, we cannot do this "exchange of values" because it may possible violate the constraint. But there may exist other transformations of values that "reflect" the point on one side of the diagonal to the other side. (see fig 2)
The "symmetry" I mentioned in the title means that if there exists solution features $x<y$, then there exists solution with $x\geq y$. How to prove it?
The claim isn't true. Let $f(x)=x^2$, $a=2$, $b=1$ $c=24$ and $z=342$. Then the equations are $2x+y=24$, $2x^2+y^2=342$, which reduces by substitution to $x^2-8x+39=0$ so $x=3, y=18$ and $x=13, y=-2$. So there is a solution with $y>x$ but none with $x>y$, because of the condition that $x, y >0$.
The original proposer's intuition fails, I think, because he has neglected the possibility that 'reflection' might take the solution outside the constrained domain.
Of course $x^2$ is only increasing for $x>0$, which is all that's relevant here. You can construct a similar counterexample with $f(x)=(x+1)^2$ for $x>0$ and $f(x)=e^x$ for $x<0$ if you care about this (you have to change 342 to 393 for the numbers to stay nice).
If you don't care about the positivity constraint, you could use $f(x)= 1 +\frac{2}{\pi} \tan^{-1} x$, with $a=2, b=1, c=\frac{1}{3}$, and $z=3$. Then the sole solution is one where $6x^3-x^2+1=0$, which only has a root where $x=-\frac{1}{2}, y=\frac{4}{3}$, so $y>x$.