Unspecified Constraint

Let $abc=k^3$, $a=kfrac{x}{y}$ and $b=kfrac{y}{z},$ where $x$, $y$ and $z$ are positives.

Thus, $$kfrac{x}{y}cdot kfrac{y}{z}cdot c=k^3,$$ which gives $$c=kfrac{z}{x}$$ and we need to prove that: $$sum_{cyc}frac{left(kfrac{x}{y}right)^3}{left(kfrac{x}{y}right)^3+k^3+left(kfrac{y}{z}right)^3}geq1$$ or $$sum_{cyc}frac{left(frac{x}{y}right)^3}{left(frac{x}{y}right)^3+1+left(frac{y}{z}right)^3}geq1$$ which says that we can assume $k=1$ or $abc=1$.

Let $x=dfrac{a}{b},, y=dfrac{b}{c},,z=dfrac{c}{a}$ then $xyz = frac ab cdot frac b c cdot frac ca =1$ and $$dfrac{a^3}{a^3+abc+b^3}=dfrac{1}{1+dfrac{bc}{a^2}+left(dfrac{b}{a}right)^3} = dfrac{1}{1+dfrac{z}{x} +left(dfrac{1}{x}right)^3} = frac{x^3}{x^3+x^2z+1} =dfrac{x^2}{x^2+zx+zy}$$ Now, using the Cauchy-Schwarz inequality, we have $$sum {dfrac{a^3}{a^3+abc+b^3}}= sum {dfrac{x^2}{x^2+zx+zy}} geqslant dfrac{(x+y+z)^2}{x^2+y^2+z^2+2(xy+yz+zx)}=1.$$