Strange inequality over the positive reals

Question

Let $a$, $b$ and $c$ be positive real numbers with $2a^3b+2b^3c+2c^3a=a^2b^2+b^2c^2+c^2a^2$. Prove that $$2ab(a-b)^2+2bc(b-c)^2+2ca(c-a)^2 \geqslant (ab+bc+ca)^2$$

Answer 1

We need to prove that $$\sum_{cyc}(2a^3b+2a^3c-5a^2b^2-2a^2bc)\geq0$$ or $$\sum_{cyc}(2a^3b+2a^3c-5a^2b^2-2a^2bc)+3\sum_{cyc}(2a^3b-a^2b^2)\geq0$$ or $$\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc)\geq0.$$

Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.

Hence, $$\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc)=$$ $$=6(u^2-uv+v^2)a^2+(5u^3+3u^2v-6uv^2+5v^3)a+uv(2u-v)^2\geq0.$$ Done!

Also, we can use C-S.

We need to prove that $$\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc)\geq0$$ or $$\sum_{cyc}(4a^3b-4a^2b^2+ab^3)\geq(a+b+c)abc$$ or $$\sum_{cyc}ab(2a-b)^2\geq(a+b+c)abc$$ or $$\sum_{cyc}\frac{(2a-b)^2}{c}\geq a+b+c.$$

Now, by C-S $$\sum_{cyc}\frac{(2a-b)^2}{c}\geq\frac{\left(\sum\limits_{cyc}(2a-b)\right)^2}{\sum\limits_{cyc}c}=a+b+c.$$