Minimum possible values of $\frac{x^2+y^2+z^2+1}{xy+yz+z}$ and $\frac{x^2+y^2+z^2+1}{xy+y+z}$ Let $k$ and $m$ be the minimum possible values of $$\frac{x^2+y^2+z^2+1}{xy+yz+z} \quad \text{and} \quad \frac{x^2+y^2+z^2+1}{xy+y+z}$$ respectively where $x,y,z$ are non-negative real numbers. What is the value of $km+k+m$?
I used the AM-GM inequality to get a minimum value for $x^2+y^2+z^2+1$ but the problem is with getting an upper bound for $xy+yz+z$ and $xy+y+z$.
This problem is from India IMC 2017 team contest.

Here is a way to use AM-GM.  In the first case, you can find $k$ if you can find suitable $\alpha, \beta$ s.t. the following AM-GMs can achieve equality simultaneously: 
$$x^2+\alpha^2 y^2 \geqslant 2\alpha x = kxy\\ (1-\alpha^2)y^2+\beta^2z^2\geqslant 2\sqrt{1-\alpha^2}\beta yz = kyz \\ (1-\beta^2)z^2+1 \geqslant 2\sqrt{1-\beta^2}z=kz$$
as summing above gets $x^2+y^2+z^2+1\geqslant k(xy+yz+z)$. Solving $2\alpha = 2\sqrt{1-\alpha^2}\beta=2\sqrt{1-\beta^2}=k$ to get $k = \sqrt5-1$, with equality when $x=1, y=z=\varphi = \frac12(\sqrt5+1)$, so we have our minimum.  
Similar approach in the second case gives $m = \sqrt5-1$ also, but this time with equality when $y=1, x=z=\varphi-1$.  Calculate $km+k+m = 4$ to finish.   
