$ x,y,z\ge 0: xy+yz+zx=3$, prove that $\frac{1}{\sqrt{3x+1}}+\frac{1}{\sqrt{3y+1}}+\frac{1}{\sqrt{3z+1}}\ge \frac{3}{2}.$

Question

Problem. If $ x,y,z\ge 0: xy+yz+zx=3$, prove that$$\frac{1}{\sqrt{3x+1}}+\frac{1}{\sqrt{3y+1}}+\frac{1}{\sqrt{3z+1}}\ge \frac{3}{2}.$$

By AM-GM, we will prove stronger:
$$\sqrt{3x+1}+\sqrt{3y+1}+\sqrt{3z+1}\ge \frac{3}{4}\sqrt{(3x+1)(3y+1)(3z+1)}.$$
I check it and it seems true. I tried to square both side, but the rest is complicated.

Hope to see more ideas. Thank you!

Answer 1

Another way.

By AM-GM $$\sum_{cyc}\frac{1}{\sqrt{1+3x}}=\sum_{cyc}\frac{4}{2\sqrt{4(1+3x)}}\geq\sum_{cyc}\frac{4}{3x+5}$$ and it's enough to prove that: $$8\sum_{cyc}(3x+5)(3y+5)\geq3\prod_{cyc}(3x+5)$$ or $$5(x+y+z)+12\geq27xyz,$$ which is true by AM-GM again.