Prove the inequality $a\frac{a+b}{a+c} + b\frac{b+c}{a+b} + c\frac{c+a}{b+c} \geq a+b+c$ for positive $a, b, c$ 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.
 A: 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,
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-a-b-c=$$
$$=\frac{a^2}{b}+\frac{b^2}{a}-a-b+\frac{b^2}{c}-c+\frac{c^2}{a}-\frac{b^2}{a}=$$
$$=\frac{(a-b)^2(a+b)}{ab}+\frac{(c-a)(c^2-b^2)}{ac}\geq0.$$
A: An AM-GM proof
By AM-GM inequality $$ab^3+abc^2\geq2\sqrt{a^2b^4c^2}=2ab^2c\tag{1}$$ Similarly we get $$bc^3+bca^2\overset{\text{AM-GM}}\geq2abc^2\tag{2}$$ and
$$ca^3+cab^2\overset{\text{AM-GM}}\geq2a^2bc\tag{3}$$
Hence adding $(1),(2),(3)$ we get the desired inequality.
