Problem 1415

Solution:
By Cauchy-Schwarz inequality and $a b c = 1$ we get
$$\begin{gathered} \sqrt{\left(b c + c a + \frac{1}{a b}\right)\left(a b + b c + \frac{1}{c a}\right)} = \sqrt{\left(b c + c a + \frac{1}{a b}\right)\left(\frac{1}{c a} + a b + b c\right)} \geq \\ \left(\sqrt{a b} \cdot \sqrt{\frac{1}{a b}} + \sqrt{b c} \cdot \sqrt{b c} + \sqrt{\frac{1}{c a}} \cdot \sqrt{c a}\right) = (2 + b c) = (2 a b c + b c) = b c(1 + 2 a) \end{gathered}$$
Analogously we get $\sqrt{\left(b c + c a + \frac{1}{a b}\right)\left(c a + a b + \frac{1}{b c}\right)} \geq c a(1 + 2 b)$ and
$\sqrt{\left(c a + a b + \frac{1}{b c}\right)\left(a b + b c + \frac{1}{c a}\right)} \geq a b(1 + 2 a)$.
Multiplying these three inequalities we get:
$$\left(a b + b c + \frac{1}{c a}\right)\left(b c + c a + \frac{1}{a b}\right)\left(c a + a b + \frac{1}{b c}\right) \geq a^{2} b^{2} c^{2}(1 + 2 a)(1 + 2 b)(1 + 2 c) =$$
$(1 + 2 a)(1 + 2 b)(1 + 2 c)$ because $a b c = 1$.

Equality holds if and only if $a = b = c = 1$.

Solution:
Using $a b c = 1$ we get
$$\begin{gathered} \left(a b + b c + \frac{1}{c a}\right)\left(b c + c a + \frac{1}{a b}\right)\left(c a + a b + \frac{1}{b c}\right) = \\ = \left(\frac{1}{c} + \frac{1}{a} + b\right)\left(\frac{1}{a} + \frac{1}{b} + c\right)\left(\frac{1}{b} + \frac{1}{c} + a\right) = \\ = \frac{(a + c + a b c)}{a c} \cdot \frac{(b + a + a b c)}{a b} \cdot \frac{(b + c + a b c)}{b c} = (a + b + 1)(b + c + 1)(c + a + 1) \end{gathered}$$
Thus, we need to prove
$$(a + b + 1)(b + c + 1)(c + a + 1) \geq (1 + 2 a)(1 + 2 b)(1 + 2 c)$$
After multiplication and using the fact $a b c = 1$ we have to prove
$$\begin{gathered} a^{2} b + a^{2} c + b^{2} c + b^{2} a + c^{2} a + c^{2} b + 3(a b + b c + c a) + 2(a + b + c) + a^{2} + b^{2} + c^{2} + 3 \geq \\ \geq 4(a b + b c + c a) + 2(a + b + c) + 9 \end{gathered}$$
So we need to prove
$$a^{2} b + a^{2} c + b^{2} c + b^{2} a + c^{2} a + c^{2} b + a^{2} + b^{2} + c^{2} \geq a b + b c + c a + 6$$
This follows from the well-known (AM-GM inequality) inequalities
$$a^{2} + b^{2} + c^{2} \geq a b + b c + c a$$
and
$$a^{2} b + a^{2} c + b^{2} c + b^{2} a + c^{2} a + c^{2} b \geq 6 a b c = 6$$