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.
 A: In fact, we can prove a stronger inequality
$$
\frac{xy+yz+zx}{x+y+z}+\sqrt{4(x^2+y^2+z^2)+6}\geq 1+3\sqrt{2} >5
$$
for all $x,y,z>0$ with $xyz=1$.
To prove the above inequality, we use the Cauchy-Schwarz Inequality to deduce that
$$
 \left( 1+1+1+\frac{3}{2}  \right)\left( 4x^2+4y^2+4z^2 +6  \right)\geq \left( 2x+2y+2z +3 \right)^2,
$$
which gives
$$ 
  \sqrt{4(x^2+y^2+z^2)+6} \geq \frac{\sqrt{2}}{3}\left( 2x+2y+2z +3 \right).
$$
As a result, we have
\begin{equation} \begin{split}
\frac{xy+yz+zx}{x+y+z}+\sqrt{4(x^2+y^2+z^2)+6} &\geq \frac{xy+yz+zx}{x+y+z}+\frac{\sqrt{2}}{3}( 2x+2y+2z)+\sqrt 2 \\
&=\frac{xy+yz+zx}{x+y+z}+ 2\sqrt 2 \left(\frac{x+y+z}{3} \right)+\sqrt 2.
\end{split}\end{equation}
Therefore, it suffices to show
$$
\frac{xy+yz+zx}{x+y+z}+ 2\sqrt 2 \left(\frac{x+y+z}{3} \right) \geq 1 +2\sqrt{2}.
$$
To do this, we note by AM-GM inequality
\begin{align}
xy + yz +zx& \geq 3 \sqrt[3]{x^2y^2z^2} =3, \\
x+y+z &\geq 3 \sqrt[3]{xyz} =3
\end{align}
Using the above inequalities, we have
\begin{equation} \begin{split}
&\frac{xy+yz+zx}{x+y+z}+ 2\sqrt 2 \left(\frac{x+y+z}{3} \right) \\
& \quad\quad\quad =\frac{xy+yz+zx}{x+y+z}+\frac{x+y+z}{3} +(2\sqrt 2-1)\left(\frac{x+y+z}{3} \right) \\
& \quad\quad\quad \geq 2 \sqrt{\frac{xy+yz+zx}{3}}+ (2\sqrt 2-1)\left(\frac{x+y+z}{3} \right) \quad (\text{by AM-GM}) \\
& \quad\quad\quad \geq 2+ 2\sqrt 2-1 \\
& \quad\quad\quad = 1+ 2\sqrt 2. 
\end{split}\end{equation}
This completes the proof.
A: Take $x=y=z=1$ and you will easily see that the right hand side is larger than $0$.
