<s>[INST] Showing $\frac{xy+yz+xz}{x+y+z}>5-\sqrt{4(x^2+y^2+z^2)+6}$ for $x,y,z>0$ and $xyz=1$ 
Let $x,y,z>0$ with $ xyz=1$ then prove that,
$$\frac{xy+yz+xz}{x+y+z}>5-\sqrt{4(x^2+y^2+z^2)+6}$$

Let $$5≤\sqrt{4(x^2+y^2+z^2)+6}\implies x^2+y^2+z^2≥\frac {25-6}{4}=\frac {19}{4}$$ then the inequality is always true.
This also shows that, if $x≥3$ or $x≥\frac {\sqrt {19}}{2}$ then the inequality is true. Since the inequality is cyclic, if $x$ or $y$ or $z≥\frac {\sqrt {19}}{2}$ then the inequality is again true.
I see that if $x,y$ is strictly increasing then $5-\sqrt{4(x^2+y^2+z^2)+6}$ is always negative.
But, I couldn't prove the general case.
 [/INST] Take $x=y=z=1$ and you will easily see that the right hand side is larger than $0$.
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