Prove $ \sum_{cyc}\frac{x}{\sqrt{x^2+8yz}} \ge 1, \forall x,y,z\gt 0 $

Question

Prove $ \sum_{cyc}\frac{x}{\sqrt{x^2+8yz}} \ge 1, \forall x,y,z\gt 0 $

I feel like the products between different variables (i.e. not x^2, y^2, z^2) give this inequality the $ \gt $ sign and I don't know what to start with. I tried setting $x+y+z= fixed$ $ value $, but that didn't work out.

Answer 1

By Holder $$\sum_{cyc}\frac{x}{\sqrt{x^2+8yz}}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{x}{\sqrt{x^2+8yz}}\right)^2\sum\limits_{cyc}x(x^2+8yz)}{\sum\limits_{cyc}x(x^2+8yz)}}\geq\sqrt{\frac{(x+y+z)^3}{\sum\limits_{cyc}x(x^2+8yz)}}.$$ Id est, it's enough to prove that $$(x+y+z)^3\geq\sum\limits_{cyc}x(x^2+8yz)$$ or $$\sum_{cyc}(x^2y+x^2z-2xyz)\geq0,$$ which is true by AM-GM.

About Holder see here:

https://math.stackexchange.com/tags/holder-inequality