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.
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.