Tricky inequality no avail to AM-GM

Question

Let $a,b,c$ be 3 distinct positive real numbers such that abc = 1. Prove that $$\frac{a^3}{\left(a-b\right)\left(a-c\right)}\ +\frac{b^3}{\left(b-c\right)\left(b-a\right)}\ +\ \frac{c^3}{\left(c-a\right)\left(c-b\right)}\ \geq 3$$

I tried AM-GM in many different ways, but it doesn't work since one of the terms on the LHS inevitably becomes negative. Any help is greatly appreciated.

Try $a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}$ so $xyz=1$ – arberavdullahu Oct 16 '16 at 20:04 — arberavdullahu, Oct 16 '16 at 20:04

score 3 · Answer 1 · answered Oct 16 '16 at 20:35

3

By AM-GM $$\sum\limits_{cyc}\frac{a^3}{(a-b)(a-c)}=-\sum\limits_{cyc}\frac{a^3}{(a-b)(c-a)}=-\sum\limits_{cyc}\frac{a^3(b-c)}{\prod\limits_{cyc}(a-b)}=a+b+c\geq3$$

answered Oct 16 '16 at 20:35

Michael Rozenberg

194,933

Could you explain how you got $- \sum\limits_{cyc}\frac{a^3(b-c)}{\prod\limits_{cyc}(a-b)}=a+b+c$ – Airdish Oct 17 '16 at 10:48
@Airdish Because $-\sum\limits_{cyc}a^3(b-c)=(a+b+c)(a-b)(b-c)(c-a)$ – Michael Rozenberg Oct 17 '16 at 11:03
Yes that is what I'd like elaborated on. Maybe how it is factorized – Airdish Oct 17 '16 at 11:17
@Airdish $-\sum\limits_{cyc}(a^3c-a^3b)=a^3c-a^3b+b^3a-c^3a+c^3b-b^3c=(b-c)(-a^3+a(b^2+bc+c^2)-bc(b+c))=...$ – Michael Rozenberg Oct 17 '16 at 11:47
@Airdish $-a^3+ab^2+abc+ac^2-b^2c-bc^2=(a-b)(-a^2-ab+bc+c^2)=(a-b)(c-a)(a+b+c)$ – Michael Rozenberg Oct 17 '16 at 12:07

Tricky inequality no avail to AM-GM

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