A difficult Vietnamese university entrance exam problem minimize/maximize $P = \frac{x}{{2x + 3y}} + \frac{y}{{y + z}}+\frac{z}{{z + x}}$? A difficult problem from the Vietnamese Entrance exam of 2011 for Natural Science and Technology group.
Note that the original problem only involve minimization.

How to minimize and maximize $P = \frac{x}{{2x + 3y}} + \frac{y}{{y + z}} + \frac{z}{{z + x}}$ when $x,y,z \in \left[ {1,4} \right]$, $x \geqslant y$ and $x \geqslant z$?

The answer to the maximization problem is $P = \frac{6}{5}$ when $x = y$.
A nice solution without Lagrange multipliers/calculus is preferred, just like the solution for the minimization problem in @Toby Mak's link.
The answer to the minimization problem is $P = \frac{{34}}{{33}}$ when $x = 4,y = 1,z = 2$ but the approach from the official government answer sheet was quite unfair.
To be particular, solving by knowing the answer before hand. It requires student to first proof something from thin air into existence
$\frac{1}{{1 + a}} + \frac{1}{{1 + b}} \geqslant \frac{2}{{1 + \sqrt {ab} }}$.

The maximization problem:
Since $P$ is homogeneous, WLOG, assume $x = 4$.
If $y = 4$, then $P = \frac{6}{5}$ for any $z$.
Indeed, $\frac65$ is the maximum of $P$, since
\begin{align*}
 P &= \frac{4}{8 + 3y} + \frac{y}{y + z} + \frac{z}{z + 4}\\
 &= \frac{6}{5} - \frac{(4 - y)[3(z - 1)(4y - z) + 25z]}{5(8 + 3y)(y + z)(z + 4)}\\
 &\le \frac{6}{5}
\end{align*}
with equality if and only if $y = 4$.
