Interesting inequality with three positive reals which seems trivial Let $a,b,c$ be positive real numbers, such that $(ab)^2 + (bc)^2 + (ca)^2 = 3$. Prove that
$(a^2 - a + 1)(b^2 - b + 1)(c^2 - c + 1) \geq 1$.
First I multiplied both sides by $(a+1)(b+1)(c+1)$ and expand the inequality:
$$a^3b^3c^3+a^3b^3+b^3c^3+c^3a^3\geq abc+ab+bc+ca$$
But stuck at this point.
After that i tried to use this inequality:
$$(a^2-a+1)(b^2-b+1) \geq \frac{a^2+b^2}{2}$$
Let $a^2=x$, $b^2=y$, and $c^2=z$.
$xy+yz+zx=3$
Then the statement is equivalent to:
$$(x+y)(y+z)(z+x) \geq 8$$

Now, since $$\prod_{cyc}(x+y)\geq\frac{8}{9}(x+y+z)(xy+xz+yz),$$ it's enough to prove that $$x+y+z\geq3,$$ which is for you.
