<s>[INST] Inequality. $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2}$ prove the following inequality: 
$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2},$$ for $a,b,c$ real positive numbers. 
Thanks :) 
 [/INST] I started this proof but then realized that there was a mistake. Below is the wrong argument. Enjoy figuring out the mistake. (The argument can be fixed though.)
\begin{align}
S(a,b,c) & = \dfrac{a^3}{a+b} + \dfrac{b^3}{b+c} + \dfrac{c^3}{c+a}\\
& = \dfrac{a^3+b^3}{a+b} + \dfrac{b^3+c^3}{b+c} + \dfrac{c^3+a^3}{c+a} - \left(\dfrac{b^3}{a+b} + \dfrac{c^3}{b+c} + \dfrac{a^3}{c+a} \right)\\
& = (a^2 - ab + b^2) + (b^2 - bc + c^2) + (c^2 - ca + a^2) - S(a,c,b)
\end{align}
Hence, $$S(a,b,c) + S(a,c,b) = (a^2 - ab + b^2) + (b^2 - bc + c^2) + (c^2 - ca + a^2) \geq ab + bc+ ca$$
If $S(a,b,c) < \dfrac{ab + bc+ ca}2$, then $S(a,c,b) < \dfrac{ac + cb+ ba}2$ which gives us
$$S(a,b,c) + S(a,c,b) < ab + bc + ca$$ Contradiction.
</s>