Help on this inequality

Question

If a,b,c are positive numbers, prove the inequality $$ \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} ≥ \frac{3}{1+abc} $$

Answer 1

Brute force works. (depending on your source, a nicer solution could exist)

Cross multiply and shift terms over, we want to show that

$$ \sum_{cyc} a^3 b^2 c^2 - 2 a^2b^2c + b^2 a + a^3 bc^2 - 2 a^2 bc + ab \geq 0 $$

This is true by AM-GM, since $ a^3b^2c+ b^2 a \geq 2 a^2 b^2 c$ and $ a^3 bc^2 + ab \geq 2 a^2bc$. Sum up the cyclic versions.

You can verify that the equality conditions imply that $a=b=c=1$.

Answer 2

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.

Answer 3

We need to prove that $$\sum_{cyc}\frac{1+abc}{a(1+b)}\geq3$$ or $$\sum_{cyc}\frac{1+abc+a+ab}{a(1+b)}\geq6$$ or $$\sum_{cyc}\frac{1+a+ab(1+c)}{a(1+b)}\geq6$$ or $$\sum_{cyc}\frac{1+a}{a(1+b)}+\sum_{cyc}\frac{b(1+c)}{1+b}\geq6,$$ which is AM-GM: $$\sum_{cyc}\frac{1+a}{a(1+b)}+\sum_{cyc}\frac{b(1+c)}{1+b}\geq3\sqrt[3]{\prod_{cyc}\frac{1+a}{a(1+b)}}+3\sqrt[3]{\prod_{cyc}\frac{b(1+c)}{1+b}}=6.$$ Done!