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!

We can use also the following way.
Full expanding gives:
$$(5a+41c)b^2+(200a^2-71ac-37c^2)b+5a^2c+41ac^2\geq0,$$ which is obvious for
$200a^2-71ac-37c^2\geq0.$
But for $200a^2-71ac-37c^2<0$ we obtain $a<c$ and it's enough to prove that
$$(200a^2-71ac-37c^2)^2-4(5a+41c)(5a^2c+41ac^2)\leq0$$ or
$$(c-a)(4a-c)(10000a^2+5375ac+1369c^2)\geq0,$$ which is true because
$$4a-c\geq4\cdot1-4=0.$$
Done!
