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 $a, b, c$

Question

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.

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 04 '17 at 12:32
@SchrodingersCat It isn’t symmetric (not the LHS), though it has cyclic symmetry. — Macavity, Oct 04 '17 at 12:37

Michael Rozenberg · Accepted Answer · 2017-10-04T13:05:29.757

2

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

edited Oct 04 '17 at 13:05

answered Oct 04 '17 at 12:41

Michael Rozenberg

194,933

Yeah, as one of variants I got this. But how we prove it? – Valentun Oct 04 '17 at 12:42
@Valentun I added something. See now. – Michael Rozenberg Oct 04 '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 04 '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 04 '17 at 12:53
what's c - s means? – Valentun Oct 04 '17 at 13:03
@Valentun It's Cauchy-Schwarz inequality. – Michael Rozenberg Oct 04 '17 at 13:07
oh, thanks, now I understand it – Valentun Oct 04 '17 at 13:10
@Valentun You are welcome! – Michael Rozenberg Oct 04 '17 at 13:17

score 0 · Answer 2 · answered Jul 30 '20 at 08:53

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.

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 $a, b, c$

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