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.
On
Let $U = [0,\infty)^3$ and $V = [0,\infty)^3 \times (0,\infty)$.
We will parameterize $U$ by elements in $V$ through following map:
$$V \in (X,Y,Z,W) \mapsto (x,y,z) = \left(\frac{X}{W},\frac{Y}{W},\frac{Z}{W}\right) \in U$$
The minimum $k$ we seek equals to
$$k = \min_{(x,y,z)\in U}\left\{\frac{x^2+y^2+z^2+1}{xy+yz+z}\right\}$$
Notice $$\frac{1}{k} = \max_{(x,y,z)\in U}\left\{\frac{xy+yz+z}{x^2+y^2+z^2+1}\right\} = \max_{(X,Y,Z,W)\in V}\left\{\frac{XY+YZ+ZW}{X^2+Y^2+Z^2+W^2}\right\}\\ = \max_{(X,Y,Z,W)\in V\cap S^3}\{ XY+YZ + ZW\} = \frac12 \max_{u \in V\cap S^3}\{ u^T\Lambda u \} $$ where $u^T = (X,Y,Z,W)$ and $\Lambda$ is the matrix $\begin{bmatrix} 0 & 1 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 0\end{bmatrix}$
Since $\Lambda$ is real symmetric, it can be diagonalized by orthogonal matrices and its eigenvalues are real. Let $\lambda_1 \ge \lambda_2 \ge \lambda_3 \ge \lambda_4$ be the eigenvalues of $\Lambda$ and $v_1, v_2, v_3, v_4$ be a corresponding set of orthonormal eigenvectors.
For any $u \in S^3$, we have $u = \sum_{k=1}^4 \alpha_k v_k$ where $\alpha_k = u^T v_k$. Furthermore,
$$u^T \Lambda u = \sum_{k=1}^4 \lambda_k \alpha_k^2 \le \lambda_1\sum_{k=1}^4 \alpha_k^2 = \lambda_1 u^T u = \lambda_1$$
This implies $$\max_{u \in V\cap S^3}\{ u^T\Lambda u \} \le \max_{u \in S^3}\{ u^T \Lambda u \} \le \lambda_1$$
Since $\Lambda$ is a non-negative irreducible, Perron-Frobenius theorem tells us $\lambda_1$ is simple and $v_1$ can be chosen to belong to $(0,\infty)^4 \subset V$. This means
$$\max_{u \in V\cap S^3}\{ u^T\Lambda u \} \ge v_1^T \Lambda v_1 = \lambda_1 v_1^T v_1 = \lambda_1$$
and hence $\max\limits_{u \in V\cap S^3}\{ u^T\Lambda u \} = \lambda_1$. As a corollary, we find
$$k = \frac{2}{\lambda_1}$$
It is not hard to work out the eigenvalues/eigenvalues of $\Lambda$. They are
$$\lambda_k = 2\cos(\frac{k\pi}{5}) \quad \text{ and } \quad v_k^T \propto \left( \sin\frac{k\pi}{5}, \sin\frac{2k\pi}{5}, \sin\frac{3k\pi}{5}, \sin\frac{4k\pi}{5}\right)$$
In particular, $\lambda_1 = 2\cos\frac{\pi}{5} = \varphi$ and $v_1^T \propto (1, \varphi, \varphi, 1)$ where $\varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio. From this, we can conclude $$k = \frac{2}{\varphi}$$ and the minimum value of $k$ is achieved at $(x,y,z) = (1,\varphi,\varphi)$.
For minimum $m$, the derivation is similar, we have
$$m = \min_{(x,y,z)\in U}\left\{\frac{x^2+y^2+z^2+1}{xy+y+z}\right\} = \frac{2}{\varphi}$$ and the minimum value is achieved at $(x,y,z) = (\frac{1}{\varphi},1,\frac{1}{\varphi})$.
Combine all these, we obtain
$$mk + m + k = \left(\frac{2}{\varphi}\right)^2 + 2 \left(\frac{2}{\varphi}\right) = \frac{4}{\varphi^2}(\varphi + 1) = 4$$
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.