The cube substitution: Σ 1/(a + b + 1) ≤ 1

Substitution. Let $x=a^{1/3}$, $y=b^{1/3}$, $z=c^{1/3}$, so $xyz=1$.

Key lemma. For $x,y>0$, $x^3+y^3-xy(x+y)=(x+y)(x-y)^2\ge0$.

Estimate. Using $1=xyz$,
$$a+b+1=x^3+y^3+xyz\ge xy(x+y)+xyz=xy(x+y+z).$$
Since $xy=\frac1z$,
$$\frac1{a+b+1}\le\frac1{xy(x+y+z)}=\frac z{x+y+z}.$$

Summation. The same holds for the other two terms, and adding gives
$$\sum\frac1{a+b+1}\le\frac{z+x+y}{x+y+z}=1.$$

Equality. Iff $x=y=z=1$.

Lesson. The constraint $abc=1$ with cubes in the lemma $x^3+y^3\ge xy(x+y)$ suggests the substitution $a=x^3$: the "1" becomes $xyz$ and the denominators become homogeneous.