If $a, b, c$ are positive real numbers such that $abc=1$ prove that $\frac{a^3}{(a-b)(a-c)} + \frac{b^3}{(b-a)(b-c)} + \frac{c^3}{(c-b)(c-a)} ≥ 3$

Question

If $a, b, c$ are distinct positive real numbers such that $abc=1$, prove that

$$\frac{a^3}{(a-b)(a-c)} + \frac{b^3}{(b-a)(b-c)} + \frac{c^3}{(c-b)(c-a)} ≥ 3.$$

I tried to do this problem by assuming that $a<b<c$. By using this, the first and the third term of the inequality are positive and the second is negative. Thus, we can obtain the minimum value of the expression by minimizing the sum of the first and the third and maximizing the second. I'm stuck at this part and would appreciate if someone could help me.

Look at the problem mentioned by Robert Z and use the indentity: $$\dfrac{a^3}{(a-b)(a-c)}+\dfrac{b^3}{(b-a)(b-c)}+\dfrac{c^3}{(c-a)(c-b)}=a+b+c$$ — Arash, Aug 15 '18 at 17:43

score 2 · Answer 1 · answered Aug 15 '18 at 18:03

As we know that $$\dfrac{a^3}{(a-b)(a-c)}+\dfrac{b^3}{(b-a)(b-c)}+\dfrac{c^3}{(c-a)(c-b)}=a+b+c$$

now apply A.M. G.M. $$\frac{a+b+c}{3}\ge (abc)^{\frac{1}{3}}$$

given $$abc=1$$ $$\frac{a+b+c}{3}\ge 1$$ $$a+b+c\ge 3$$

replcae $a+b+c$ with your expression.

$$\dfrac{a^3}{(a-b)(a-c)}+\dfrac{b^3}{(b-a)(b-c)}+\dfrac{c^3}{(c-a)(c-b)}\ge 3$$

If $a, b, c$ are positive real numbers such that $abc=1$ prove that $\frac{a^3}{(a-b)(a-c)} + \frac{b^3}{(b-a)(b-c)} + \frac{c^3}{(c-b)(c-a)} ≥ 3$

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