Proving $\frac{a}{a+ b}+ \frac{b}{b+ c}+ \frac{c}{2c+ 5a}\geq \frac{38}{39}$ $$a, b, c \in \left [ 1, 4 \right ] \text{ }
\frac{a}{a+ b}+ \frac{b}{b+ c}+ \frac{c}{2c+ 5a}\geq \frac{38}{39}$$
This is an [old problem of mine in AoPS]
(https://artofproblemsolving.com/community/u372289h1606772p10020940).
First solution
$$ab\geq  1\Leftrightarrow  \frac{1}{1+ a}+ \frac{1}{1+ b}- \frac{2}{1+ \sqrt{ab}}= \frac{\left ( \sqrt{ab}- 1 \right )\left ( \sqrt{a}- \sqrt{b} \right )^{2}}{\left ( 1+ a \right )\left ( 1+ b \right )\left ( 1+ \sqrt{ab} \right )}\geq  0$$
then
$$\frac{1}{1+ a}+ \frac{1}{1+ b}\geq \frac{2}{1+ \sqrt{ab}}$$
Thus, we have
$$P= \frac{1}{1+ \frac{b}{a}}+ \frac{1}{1+ \frac{c}{b}}+ \frac{c}{2c+ 5a}\geq  \frac{2}{1+ \sqrt{\frac{c}{a}}}+ \frac{\frac{c}{a}}{2\frac{c}{a}+ 5}\geq  \frac{38}{39}$$
which is true by $\frac{c}{a}\leq 4$
How about another solution? I hope to see more. Thanks!

You want to minimize the smooth function $F(a,b,c) = {\frac {a}{a+b}}+{\frac {b}{b+c}}+{\frac {c}{2\,c+5\,a}}$ over the cube $1 \le a,b,c \le 4$.  The candidates for minimizer are critical points in the interior, points on a face where the gradient is orthogonal to the face, points on an edge where the gradient is orthogonal to the edge, and vertices.  It's a bit tedious, but quite routine.  The minimum value turns out to be $38/39$, achieved at $(a,b,c) = (1,2,4)$.
