<s>[INST] For $x,y,z∈ℝ^{+}$,without using Hölder's inequality prove that $\frac{x}{\sqrt{x^2+8yz}}+\frac{y}{\sqrt{y^2+8xz}}+\frac{z}{\sqrt{z^2+8xy}}\geq1$. 
For $x,y,z∈ℝ^{+}$, prove that $\frac{x}{\sqrt{x^2+8yz}}+\frac{y}{\sqrt{y^2+8xz}}+\frac{z}{\sqrt{z^2+8xy}}\geq1$.

In this question solution used Hölder's inequality, but I am looking a solution without using Hölder's inequality.
I first tried using AM-GM inequality but not results clearly.
Then I used Cauchy-Schwarz inequality which gave,
$${(\sum_{cyc}x²)(\sum_{cyc}\frac{1}{x²+8yz})\geq(\sum_{cyc}\frac{x}{\sqrt{x²+8yz}})}^2$$
But I couldn't solve it. Can anyone help me out?
 [/INST] Another way.
By AM-GM, C-S and AM-GM again we obtain:$$\sum_{cyc}\frac{x}{\sqrt{x^2+8yz}}=\sum_{cyc}\frac{2x(x+y+z)}{2\sqrt{(x+y+z)^2(x^2+8yz)}}\geq\sum_{cyc}\frac{2x(x+y+z)}{(x+y+z)^2+x^2+8yz}=$$
$$=\sum_{cyc}\frac{2x^2(x+y+z)}{x(x+y+z)^2+x^3+8xyz}\geq\frac{2(x+y+z)^3}{\sum\limits_{cyc}(x(x+y+z)^2+x^3+8xyz}=$$
$$=\frac{2(x+y+z)^3}{(x+y+z)^3+x^3+y^3+z^3+24xyz}\geq1.$$
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