# prove the inequality for positive a, b, c

I faced with 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:

$bc^3 + a^3 c + a b^3 \geq a b^2 c + a^2 b c + a b c^2$

Then I tried to prove it using MMI, but it gave me nothing.

Please just give me a hint, maybe I miss something important.

• Use the fact that the inequality is symmetric in a, b, c. That is, replace a by b, b by c and c by a... inequality remains the same. – SchrodingersCat Oct 4 '17 at 12:32
• @SchrodingersCat It isn’t symmetric (not the LHS), though it has cyclic symmetry. – Macavity Oct 4 '17 at 12:37

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, $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-a-b-c=$$ $$=\frac{a^2}{b}+\frac{b^2}{a}-a-b+\frac{b^2}{c}-c+\frac{c^2}{a}-\frac{b^2}{a}=$$ $$=\frac{(a-b)^2(a+b)}{ab}+\frac{(c-a)(c^2-b^2)}{ac}\geq0.$$

• Yeah, as one of variants I got this. But how we prove it? – Valentun Oct 4 '17 at 12:42
• @Valentun I added something. See now. – Michael Rozenberg Oct 4 '17 at 12:51
• can you explain why we can say that a^2/b.... >= a^2/a....? May be using some theorem? – Valentun Oct 4 '17 at 12:51
• @Valentun I used Rearrangement. I can prove it by another way: C-S, AM-GM, SOS, SS, BW. Choose. – Michael Rozenberg Oct 4 '17 at 12:53
• what's c - s means? – Valentun Oct 4 '17 at 13:03