I faced problem proving this inequality for positive $a$, $b$, $c$:
$a\frac{a+b}{a+c} + b\frac{b+c}{a+b} + c\frac{c+a}{b+c} \geq a+b+c$
I tried to simplify it and I got that:
$bc^3 + a^3 c + a b^3 \geq a b^2 c + a^2 bc + a b c^2$
Then I tried to prove it using MMI, but it gave me nothing.
Please just give me a hint, maybe I am missing something important.
We need to prove that $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq a+b+c,$$ which is true by Rearrangement because $(a^2,b^2,c^2)$ and $\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$ are opposite ordered,
which gives $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{a^2}{a}+\frac{b^2}{b}+\frac{c^2}{c}=a+b+c.$$
C-S also works: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{(a+b+c)^2}{a+b+c}=a+b+c.$$
AM-GM: $$\sum_{cyc}\left(\frac{a^2}{b}+b\right)\geq2\sum_{cyc}\left(\frac{a^2}{b}\cdot b\right)=2\sum_{cyc}a.$$ SS:
Let $c=\min\{a,b,c\}$.
Thus, $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-a-b-c=$$ $$=\frac{a^2}{b}+\frac{b^2}{a}-a-b+\frac{b^2}{c}-c+\frac{c^2}{a}-\frac{b^2}{a}=$$ $$=\frac{(a-b)^2(a+b)}{ab}+\frac{(c-a)(c^2-b^2)}{ac}\geq0.$$