Problem
Let $a,b,c\ge 0$,and such $ab+bc+ca=3$, show that $$\sum_{cyc}a^2b^2+\sum_{cyc}\dfrac{12a^2b^2c^2}{(a+b)^2}\ge 12abc\tag{1}$$
A few hours ago, I asked for an error inequality.
Wrong inequality:Let $a,b,c\ge 0$,and such $ab+bc+ca=3$, show that $$\sum_{cyc}a^2b^2+\sum_{cyc}\dfrac{12a^2b^2c^2}{(a+b)^2}\ge 12$$
Now,I think inequality $(1)$ is right,I want use well know Iran 96 inequality $$\sum_{cyc}\dfrac{1}{(a+b)^2}\ge\dfrac{9}{4(ab+bc+ac)}=\dfrac{3}{4}$$It suffices to prove that $$\sum_{cyc}a^2b^2+9a^2b^2c^2\ge 12abc?$$
Since for $abc=0$ our inequality is obvious, we can assume that $abc\neq0.$
Let $bc=x$, $ac=y$ and $ab=z$.
Thus, $x+y+z=3$, $a=\sqrt{\frac{yz}{x}}$, $b=\sqrt{\frac{xz}{y}},$ $c=\sqrt{\frac{xy}{z}}$ and we need to prove that $$x^2+y^2+z^2+12xyz\sum_{cyc}\frac{1}{\left(\sqrt{\frac{yz}{x}}+\sqrt{\frac{xz}{y}}\right)^2}\geq4\sqrt{3xyz(x+y+z)}$$ or $$\sum_{cyc}\left(x^2+\frac{12x^2y^2}{(x+y)^2}\right)\geq4\sqrt{3xyz(x+y+z)}$$ or $$\sum_{cyc}(x^2-xy)+4(xy+xz+yz)-4\sqrt{3xyz(x+y+z)}\geq3\sum_{cyc}\left(xy-\frac{4x^2y^2}{(x+y)^2}\right)$$ or $$\frac{1}{2}\sum_{cyc}(x-y)^2+\frac{2\sum\limits_{cyc}z^2(x-y)^2}{xy+xz+yz+\sqrt{3xyz(x+y+z)}}\geq3\sum_{cyc}\frac{xy(x-y)^2}{(x+y)^2}$$ or $$\sum_{cyc}(x-y)^2\left(1+\frac{4z^2}{xy+xz+yz+\sqrt{3xyz(x+y+z)}}-\frac{6xy}{(x+y)^2}\right)\geq0$$ and since $xy+xz+yz\geq\sqrt{3xyz(x+y+z)},$ it's enough to prove that $$\sum_{cyc}(x-y)^2\left(1+\frac{2z^2}{xy+xz+yz}-\frac{6xy}{(x+y)^2}\right)\geq0.$$ Now, let $x\geq y\geq z$.
Thus, by AM-GM $$1+\frac{2x^2}{xy+xz+yz}-\frac{6yz}{(y+z)^2}\geq1+\frac{2x^2}{xy+xz+yz}-\frac{6yz}{4yz}=\frac{4x^2-xy-xz-yz}{2(xy+xz+yz)}\geq0.$$ Also, by AM-GM again we have $$1+\frac{2y^2}{xy+xz+yz}-\frac{6xz}{(x+z)^2}=$$ $$=\frac{2(x+z)^2y^2+(x^3-3x^2z-3xz^2+z^3)y+xz(x^2-4xz+z^2)}{(xy+xz+yz)(x+z)^2}=$$ $$=\frac{y(2(x+z)^2y+x^3-3x^2z-3xz^2+z^3)+xz(x^2-4xz+z^2)}{(xy+xz+yz)(x+z)^2}\geq$$ $$\geq\frac{y(2(x+z)^2z+x^3-3x^2z-3xz^2+z^3)+xz(x^2-4xz+z^2)}{(xy+xz+yz)(x+z)^2}=$$ $$=\frac{y(x^3-x^2z+xz^2+3z^3)+xz(x^2-4xz+z^2)}{(xy+xz+yz)(x+z)^2}\geq$$ $$\geq\frac{z(x^3-x^2z+xz^2+3z^3)+xz(x^2-4xz+z^2)}{(xy+xz+yz)(x+z)^2}=$$ $$=\frac{z(2x^3-5x^2z+2xz^2+3z^3)}{(xy+xz+yz)(x+z)^2}=\frac{z\left(3\cdot\frac{2}{3}x^3+2xz^2+6\cdot\frac{1}{2}z^3-5x^2z\right)}{(xy+xz+yz)(x+z)^2}\geq$$ $$\geq\frac{x^2z^2\left(10\sqrt[10]{2\left(\frac{2}{3}\right)^3\left(\frac{1}{2}\right)^6}-5\right)}{(xy+xz+yz)(x+z)^2}\geq0.$$ Id est, $$\sum_{cyc}(x-y)^2\left(1+\frac{2z^2}{xy+xz+yz}-\frac{6xy}{(x+y)^2}\right)\geq$$ $$\geq(x-z)^2\left(1+\frac{2y^2}{xy+xz+yz}-\frac{6xz}{(x+z)^2}\right)+(x-y)^2\left(1+\frac{2z^2}{xy+xz+yz}-\frac{6xy}{(x+y)^2}\right)\geq$$ $$\geq(x-y)^2\left(1+\frac{2y^2}{xy+xz+yz}-\frac{6xz}{(x+z)^2}\right)+(x-y)^2\left(1+\frac{2z^2}{xy+xz+yz}-\frac{6xy}{(x+y)^2}\right)=$$ $$=2(x-y)^2\left(1+\frac{y^2+z^2}{xy+xz+yz}-\frac{3xz}{(x+z)^2}-\frac{3xy}{(x+y)^2}\right)$$ and it's enough to prove that $$1+\frac{y^2+z^2}{xy+xz+yz}-\frac{3xz}{(x+z)^2}-\frac{3xy}{(x+y)^2}\geq0$$ or $$(y+z)x^5-yzx^4+3(y^3-2y^2z-2yz^2+z^3)x^3+(y^4+4y^3z-8y^2z^2+4yz^3+z^4)x^2+$$ $$+2yz(y^3+y^2z+yz^2+z^3)x+y^2z^2(y^2+yz+z^2)\geq0.$$ Now, let $x=t\sqrt{yz}$ and since $$y+z\geq2\sqrt{yz},$$ $$y^3-2y^2z-2yz^2+z^3\geq-2\sqrt{y^3z^3},$$ $$y^4+4y^3z-8y^2z^2+4yz^3+z^4\geq2y^2z^2,$$ $$2yz(y^3+y^2z+yz^2+z^3)\geq8\sqrt{y^5z^5}$$ and $$y^2z^2(y^2+yz+z^2)\geq3y^3z^3,$$ it's enough to prove that $$2t^5-t^4-6t^3+2t^2+8t+3\geq0$$ or $$(t+1)^2(2t^3-5t^2+2t+3)\geq0,$$ which we proved already by AM-GM.
Done!