Problem 1663

a) The triple $(a, b, c) = (0.01, 1000, 1000)$ does not satisfy the inequality.

b) We will prove that the inequality holds true for all positive real numbers $a, b$, and $c$. Denote by $x = \frac{a}{b}$, $y = \frac{b}{c}$, and $z = \frac{c}{a}$ - then $xyz = 1$ and $(xy + y)^2 + (yz + z)^2 + (zx + x)^2 \ge x + y + z + 9$. This is equivalent to $x^2 + y^2 + z^2 + 2(xy^2 + yz^2 + zx^2) + x^2y^2 + y^2z^2 + z^2x^2 \ge x + y + z + 9$. Applying AM-GM, we get

$xy^2 + yz^2 + zx^2 \ge 3\sqrt[3]{x^3y^3z^3} = 3$ and $x^2y^2 + y^2z^2 + z^2x^2 \ge 3\sqrt[3]{x^4y^4z^4} = 3$. Thus, it suffices to prove that $x^2 + y^2 + z^2 \ge x + y + z$. But $x^2 + y^2 + z^2 \ge \frac{(x+y+z)^2}{3}$ (it is equivalent to $(x-y)^2 + (y-z)^2 + (z-x)^2 \ge 0$), so we need $x + y + z \ge 3$, which follows from the AM-GM inequality $x + y + z \ge 3\sqrt{xyz} = 3$.