Proving $\frac{a}{a+ b}+ \frac{b}{b+ c}+ \frac{c}{2c+ 5a}\geq \frac{38}{39}$

Question

$$a, b, c \in \left [ 1, 4 \right ] \text{ } \frac{a}{a+ b}+ \frac{b}{b+ c}+ \frac{c}{2c+ 5a}\geq \frac{38}{39}$$ 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!

Answer 1

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!

Answer 2

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 $(a,b,c) = (1,2,4)$.