Problem 1185

Claim. For any positive real numbers $x, y$ with $xy \ge 1$, we have
$$(x^2 + 1)(y^2 + 1) \ge \left(\left(\frac{x+y}{2}\right)^2 + 1\right)^2. \qquad (1)$$
Proof. Note that $xy \ge 1$ implies
$$\left(\frac{x+y}{2}\right)^2 - 1 \ge xy - 1 \ge 0.$$
We find that
$$\begin{aligned} (x^2 + 1)(y^2 + 1) &= (xy - 1)^2 + (x + y)^2 \\ &\le \left(\left(\frac{x+y}{2}\right)^2 - 1\right)^2 + (x+y)^2 \\ &= \left(\left(\frac{x+y}{2}\right)^2 + 1\right) \\ &\le \left(\left(\frac{x+y}{2}\right) + 1\right)^2 \end{aligned}$$
Without loss of generality, assume $a \ge b \ge c$. This implies $a \ge 1$.
Let
$$d = \frac{a+b+c}{3}.$$
Note that
$$ad = \frac{a(a+b+c)}{3} \ge \frac{1+1+1}{3} = 1.$$
Then we can apply Eq. (1) to the pair $(b, c)$. We get
$$(a^2 + 1)(b^2 + 1)(c^2 + 1)(d^2 + 1) \le \left( \left( \frac{a+d}{2} + 1 \right)^2 + \left( \frac{b+c}{2} + 1 \right)^2 \right) \quad (2)$$
Next, from
$$\frac{a+d}{2} \cdot \frac{b+c}{2} \ge \sqrt{ad} \cdot \sqrt{bc} \ge 1,$$
we can apply Eq. (1) again to the pair $(\frac{a+d}{2}, \frac{b+c}{2})$. Together with Eq. (2), we have
$$\begin{aligned} (a^2 + 1)(b^2 + 1)(c^2 + 1)(d^2 + 1) &\le \left( \left( \frac{a+b+c+d}{4} \right)^2 + 1 \right)^4 \\ &= (d^2 + 1)^4. \end{aligned}$$
Therefore, $(a^2+1)(b^2+1)(c^2+1) \le (d^2+1)^3$, and the desired inequality follows by taking cube root of both sides.