<s>[INST] Help on this inequality If a,b,c are positive numbers, prove the inequality
$$
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} ≥ \frac{3}{1+abc}
$$
 [/INST] Let $g=\sqrt[3]{abc}$, and let $a=gx, b=gy, c=gz$, so that $xyz=1$. Then the inequality can be rewritten as:
$$\sum_{cyc}\frac{1}{gx(1+gy)}\ge \frac{1}{1+g^3}$$
Now we utilize the substitution $x=\frac{v}{u}, y=\frac{w}{v}, z=\frac{u}{w}$ for reals $u, v, w$. (Note: such a substitution always exists, take for example $u=1, v=x, w=xy$.) Then the inequality becomes:
$$\sum_{cyc}\frac{1}{g\frac{v}{u}+g^2\frac{w}{u}}\ge \frac{1}{1+g^3}$$
$$\sum_{cyc}\frac{u}{vg+wg^2}\ge \frac{1}{1+g^3}$$
But this is true because:
$$\sum_{cyc}=\frac{u}{vg+wg^2}=\sum_{cyc}\frac{u^2}{uvg+uwg^2}\ge \frac{(u+v+w)^2}{(uv+vw+wu)(g+g^2)}\ge \frac{3}{g+g^2}\ge \frac{3}{1+g^3}$$
And so the inequality is proven. 
Note: This idea of transforming a nonhomogeneous inequality into a homogeneous one is pretty useful for this type of inequality.
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