inequality under condition x+y+z＝18xyz

Question

$x,y,z \gt 0 \qquad x+y+z＝18xyz$

prove： $$\sum_{cyc}\frac{x}{\sqrt{x^2+2yz+1}}\geq 1$$

I can't find other example of $x+y+z=18xyz$

Answer 1

Let $\frac{1}{6xy}=c$, $\frac{1}{6xz}=b$ and $\frac{1}{6yz}=a$.

Hence, $a+b+c=3$ and we need to prove that $\sum\limits_{cyc}\frac{\sqrt{\frac{a}{6bc}}}{\sqrt{\frac{a}{6bc}+2\sqrt{\frac{b}{6ac}\cdot\frac{c}{6ab}}+1}}\geq1$ or $$\sum\limits_{cyc}\sqrt{\frac{a^2}{a^2+2bc+6abc}}\geq1$$ By Holder $$\left(\sum\limits_{cyc}\sqrt{\frac{a^2}{a^2+2bc+6abc}}\right)^2\sum\limits_{cyc}a(a^2+2bc+6abc)\geq(a+b+c)^3$$ Id est, it remains to prove that $$(a+b+c)^3\geq\sum\limits_{cyc}a(a^2+2bc+6abc)$$ or $$(a+b+c)^3\geq\sum\limits_{cyc}(a^3+2abc+6a^2bc)$$ or $$(a+b+c)^3\geq\sum\limits_{cyc}(a^3+8abc)$$ or $$\sum\limits_{cyc}c(a-b)^2\geq0$$.

Done!