The given condition is equivalent to $1+(abc)^2=(ab)^2+(bc)^2+(ca)^2+abc(a^2b+b^2c+c^2a)$. Using AM-GM we get $1+(abc)^2\ge 3(abc)^{4/3}+3(abc)^2$ and therefore $2x^6+3x^4-1\le 0$, where $x=(abc)^{1/3}$. But $2x^6+3x^4-1=(2x^2-1)(x^2+1)^2$ and thus $abc\le \frac{1}{2\sqrt{2}}$. Now $(ab)^3+(bc)^3+(ca)^3\ge 3(abc)^2\ge 6\sqrt{2}(abc)^3$ and we are done.

Nice. Thanks. If \(a,b,c\) be positive real numbers such that \(\dfrac{ab}{1+bc}+\dfrac{bc}{1+ca}+\dfrac{ca}{1+ab}=1\). Then$$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \ge 6 .$$

tk1 wrote: The given condition is equivalent to $1+(abc)^2=(ab)^2+(bc)^2+(ca)^2+abc(a^2b+b^2c+c^2a)$. Using AM-GM we get $1+(abc)^2\ge 3(abc)^{4/3}+3(abc)^2$ and therefore $2x^6+3x^4-1\le 0$, where $x=(abc)^{1/3}$. But $2x^6+3x^4-1=(2x^2-1)(x^2+1)^2$ and thus $abc\le \frac{1}{2\sqrt{2}}$. Now $(ab)^3+(bc)^3+(ca)^3\ge 3(abc)^2\ge 6\sqrt{2}(abc)^3$ and we are done. Nice. Thanks.

luofangxiang wrote: they are $x^2+y^2+z^2\ge \frac{3}{4}$ and $\frac{3}{2}\ge x+y+z$ where $\sum_{cyc} \frac{x}{1+y}=1$ http://artofproblemsolving.com/community/c6h1318217p7091717 http://artofproblemsolving.com/community/c6h1318217p7097702

sqing wrote: Let \(a,b,c\) be positive real numbers such that \(\dfrac{ab}{1+bc}+\dfrac{bc}{1+ca}+\dfrac{ca}{1+ab}=1\). Prove that $$\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3} \ge 6\sqrt{2}$$ $$\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3} \ge \frac{3\sqrt{2}}{2}(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})-3\sqrt{2}\ge 6\sqrt{2}$$

Let $ab=x, bc=y, ca=z\implies a=\sqrt{\frac{xz}{y}}, b=\sqrt{\frac{xy}{z}}, z=\sqrt{\frac{yz}{x}}$. Also, we have $\frac{x}{1+y}+\frac{y}{1+z}+\frac{z}{1+x}=1$. Also, $$\sum\frac{1}{a^3}=\sum\frac{y^3}{xyz\sqrt{xyz}}\geq \frac{3}{\sqrt{xyz}}\geq 6\sqrt{2}$$where the last inequality holds from P2 of Mumbai zone RMO of the same year.

file:///C:/Users/arjun/OneDrive/0_Arjun/IMO/RMO%20past%20year%20papers/my%20solution,%20RMO%202016-2%20(paper%204).pdf Go to the above link to check my solution to the problem (just copy-paste this link). I used Engel's form of cauchy schwarz inequality (also known as 'titu's lemma') to solve thus problem. Personally, I think that this is the fastest possible way to solve this problem.