Let $a,b,c$ be positive real numbers such that $\min(ab,bc,ca) \ge 1$. Prove that $\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le \left(\frac{a+b+c}{3}\right)^2 + 1$.
I tried to use the AM-GM inequality but that gives $\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le \frac{(a+b+c)^2}{3}$. I also proved the inequality but with 2 variables. $(a^2+1)(b^2+1) \le \left( \left( \frac{a+b}{2} \right)^2+1 \right)^2$ with $ab \ge 1$. If necessary I can put the proof of that here.
I suppose you already proved that $$(a^2+1)(b^2+1) \le \left( \left(\frac{a+b}{2} \right) ^2+1 \right)^2 \tag{1}$$ Denote $(x,y,z) = \left(\frac{a+b}{2},\frac{b+c}{2},\frac{c+a}{2} \right)$, from $(1)$, it's easy to deduce that $$LHS = \sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le \sqrt[3]{\left( x ^2+1 \right)\left( y ^2+1 \right)\left( z ^2+1 \right)} \tag{2}$$
From $(2)$, by taking the logarithm of both sides, the inequality in the question is equivalent to $$ \frac{1}{3} \left(\ln (x^2+1) +\ln (y^2+1)+\ln (z^2+1) \right) \le \ln \left(\left(\frac{(x+y+z}{3}\right)^2+1 \right) \tag{3}$$
From the assumption $ \min{\{ab,bc,ca\}} \ge 1$, we deduce that $x,y,z \ge 1$ (it suffices to apply the AM-GM inequality $x = (a+b)/2 \ge \sqrt{ab} \ge 1$).
Applying the Jensen's inequality to the function $f(t):= \ln(t^2 +1)$ which is concave for $t \ge 1$ (indeed, $f''(t) = 2\frac{1-t^2}{(t^2+1)^2} \le 0$ for $t \ge 1$), then the inequality $(3)$ holds true.
The equality occurs if and only if $x = y= z$ or $a = b= c$.
