If $xyz = x+y+z$ for $x,y,z > 0$, then $\sqrt{1+x^2} + \sqrt{1+y^2} + \sqrt{1+z^2} \ge 6$

Question

If $x,y,z \in \Bbb R_{>0}$ satisfy $xyz = x+y+z$, prove that $$\sqrt{1+x^2} + \sqrt{1+y^2} + \sqrt{1+z^2} \ge 6$$

We can express $1+x^2$ as $$1+x^2 = (1-xy)(1-xz) = \frac{(y-xyz)(z-xyz)}{yz} = \frac{(x+z)(x+y)}{yz}$$ since $x + y + z = xyz$ implies $1+ x^2 + xy + xz = 1 + x^2yz$. Thus, our desired inequality boils down to $$\sqrt{\frac{(x+z)(x+y)}{yz}} + \sqrt{\frac{(y+z)(y+x)}{xz}} + \sqrt{\frac{(z+x)(z+y)}{xy}} \ge 6$$ which looks more complicated. Perhaps I am not headed in the right direction? I'd appreciate any hints or solutions. Thank you!

Answer 1

Your idea gives a simple solution.

Indeed, by AM-GM twice we obtain: $$\sum_{cyc}\sqrt{\frac{(x+y)(x+z)}{yz}}\geq3\sqrt[3]{\frac{\prod\limits_{cyc}(x+y)}{xyz}}\geq3\sqrt[3]{\frac{8xyz}{xyz}}=6.$$

Answer 2

We have: $$xyz = x+y+z \ge 3\sqrt[3]{xyz} \Longrightarrow xyz \ge 3^{3/2}$$ and for $a\in \{x,y,z \}$: $$\sqrt{1+a^2} = \sqrt{1+\frac{a^2}{3}+\frac{a^2}{3}+\frac{a^2}{3}}\ge\sqrt{4\sqrt[4]{\frac{a^6}{3^3}}}=\frac{2}{3^{3/8}}\cdot a^{3/4}$$ Then $$\begin{align} LHS &\ge \frac{2}{3^{3/8}}\cdot \left(x^{3/4}+y^{3/4}+z^{3/4} \right) \ge \frac{2}{3^{3/8}}\cdot3\cdot\sqrt[3]{(xyz)^{3/4}}= \frac{6}{3^{3/8}}(xyz)^{1/4}\\ & \ge \frac{6}{3^{3/8}} \cdot 3^{\frac{3}{2}\cdot \frac{1}{4}} = 6 \end{align}$$ The equality occurs when $ x=y=z= \sqrt{3}$.

Q.E.D