Question:
Find all functions $f$ from the positive real numbers to the positive real numbers for which $f(x)\le f(y)$ whenever $x\le y$ and $$ f(x^4)+f(x^2)+f(x)+f(1)=x^4+x^2+x+1 $$ for all $x>0$.
My progress:
Substituting $x=1$ gives $4f(1)=4$, so $f(1)=1$, and $f(x)\le 1$ when $x<1$ and $f(x)\ge 1$ when $x>1$. This also means that $f(x^4)+f(x^2)+f(x)=x^4+x^2+x$. Substituting $x=x^2$ gives $f(x^8)+f(x^4)+f(x^2)=x^8+x^4+x^2$, and so $f(x)-x=f(x^8)-x^8$, and knowing the value of $f(x)$ provides the values of $f\left(x^{8^n}\right)$ where $n\in\mathbb{Z}$. I couldn't get any further than this in my attempt at the question.