Let $ a,b,c$ be the side-lengths of a triangle. Prove that: $$ \frac {a(b + c)}{a^2 + bc} + \frac {b(c + a)}{b^2 + ca} + \frac {c(a + b)}{c^2 + ab}\leq 3 $$$$\frac{a(b+c)}{a^2+2bc}+\frac{b(c+a)}{b^2+2ca}+\frac{c(a+b)}{c^2+2ab} \leq 2$$

This problem was proposed by Mudok (probably)

???? Can you prove it

still no solution, bump

I will prove it for triangle sides $a,b,c$ and same thing maybe works for the main problem, I am not sure. $\sum \frac{(a-b)(a-c)}{a^2+bc} \ge 0 \iff \sum (b-c)^2(2ac+2ac-bc-a^2)(a^2+bc) \ge 0 \leftarrow \sum (b-c)^2(ab+ac-bc)(a^2+bc) \overset{*} \ge 0$. Assume that $a\ge b\ge c$. $$(a-b)^2(ac+bc-ab)(c^2+ab)+(a-c)^2(bc+ba-ac) \ge (a-b)^2(bc^3+b^3c+a(b^3+c^3)-a^2(b-c)^2)=$$$$(a-b)^2(bc^3+b^3c+a(b+c)(b^2+c^2-bc)-a^2(b-c)^2)\ge 0$$ Thus $*$ is true.

WLOG $a \geq b \geq c$ and it's sufficient to show that \[\frac{ba + bc}{b^2 + ac} - 1 \leq 1 - \frac{ab + ac}{a^2 + bc} + 1 - \frac{ca + cb}{c^2 + ab} \quad \Leftrightarrow\]\[\frac{(a - b)(b - c)}{b^2 + ac} \leq \frac{(a - b)(a - c)}{a^2 + bc} + \frac{(a - c)(b - c)}{c^2 + ab}. \eqno (1)\]If $a = b$ or $b = c$ then $(1)$ is obviously true. Now we assume $a > b > c$. \[\frac{1}{(a - c)(b^2 + ac)} \leq \frac{1}{(b - c)(a^2 + bc)} + \frac{1}{(a - b)(c^2 + ab)}.\]\[\frac{1^2}{(b - c)(a^2+bc)} + \frac{1^2}{(a - b)(c^2 + ab)} \geq \frac{(1+1)^2}{(b - c)(a^2+bc) + (a - b)(c^2+ab)} = \frac{4}{(a - c)(2ab+2bc - b^2 - ac)}.\]\[\frac{4}{(a - c)(2ab+2bc-b^2-ac)}\geq \frac{1}{(a-c)(b^2+ac)} \quad \Leftrightarrow \]\[5b^2+5ac \geq 2(ab+bc) \quad \Leftrightarrow \quad \frac{5}{2} \geq \frac{b(c+a)}{b^2+ac},\]which is true by condition.