Unspecified Constraint

Question

In the following question-

Let a, b, c be positive real numbers. Prove that $$\sum_{cyc} {a^3\over a^3+b^3+abc} \ge 1.$$

In here, there is no constraint given. But in the solution, the author assumes cyclic substitutions of $x={b\over a}$.
But that means xyz = 1 How can this happen if no constraint is given? Thanks!

Answer 1

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

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

Answer 2

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 $$\sum {\dfrac{a^3}{a^3+abc+b^3}}= \sum {\dfrac{x^2}{x^2+zx+zy}} \geqslant \dfrac{(x+y+z)^2}{x^2+y^2+z^2+2(xy+yz+zx)}=1.$$