Prove inequality $\sum\frac{{{a^3} - {b^3}}}{{{{\left( {a - b} \right)}^3}}}\ge \frac{9}{4}$ Given $a,b,c$ are positive number. Prove that $$\frac{a^3-b^3}{\left(a-b\right)^3}+\frac{b^3-c^3}{\left(b-c\right)^3}+\frac{c^3-a^3}{\left(c-a\right)^3}\ge \frac{9}{4}$$

$$\Leftrightarrow \sum \frac{3(a+b)^2+(a-b)^2}{(a-b)^2} \ge 9$$
$$\Leftrightarrow \frac{(a+b)^2}{(a-b)^2}+\frac{(b+c)^2}{(b-c)^2}+ \frac{(c+a)^2}{(c-a)^2}\ge 2$$
Which $$\frac{a+b}{a-b}.\frac{b+c}{b-c}+\frac{b+c}{b-c}.\frac{c+a}{c-a}+\frac{c+a}{c-a}.\frac{a+b}{a-b} =-1$$
Use $x^2+y^2+z^2 \ge -2(xy+yz+zx)$ then inequality right
I don't know why use $x^2+y^2+z^2 \ge -2(xy+yz+zx)$ then inequality right?
P/s: Sorry i knowed, let $x=\frac{a+b}{a-b}$
We have $(x+1)(y+1)(z+1)=(x-1)(y-1)(z-1) => xy+yz+zx=-1 $
 A: I think it means that the variables are reals.
By your identity we obtain $$\left(\sum_{cyc}\frac{a+b}{a-b}\right)^2=\sum_{cyc}\frac{(a+b)^2}{(a-b)^2}-2\geq0$$
Thus, $$\sum_{cyc}\frac{(a+b)^2}{(a-b)^2}\geq2$$ or
$$\sum_{cyc}\left(\frac{(a+b)^2}{(a-b)^2}-1\right)\geq-1$$ or
$$\sum_{cyc}\frac{4ab}{(a-b)^2}\geq-1$$ or
$$\sum_{cyc}\frac{3ab}{(a-b)^2}\geq-\frac{3}{4}$$ or
$$\sum_{cyc}\left(\frac{3ab}{(a-b)^2}+1\right)\geq3-\frac{3}{4}$$ or
$$\sum_{cyc}\frac{a^2+ab+b^2}{(a-b)^2}\geq\frac{9}{4}$$ or
$$\sum_{cyc}\frac{a^3+b^3}{(a-b)^3}\geq\frac{9}{4}.$$
Done!
A: $$a^2-2ab+b^2\le a^2+ab+b^2$$
$$1 \le \frac{a^2+ab+b^2}{(a-b)^2}$$$$1\le\frac{a^3-b^3}{(a-b)^3}$$$$\frac{9}{4}<3\le\sum_{cyc}\frac{a^3-b^3}{(a-b)^3}$$
A: $$(x+y+z)^2\ge 0\\x^2+y^2+z^2+2xy+2xz+2yz\ge 0\\x^2+y^2+z^2\ge -2(xy+yz+zx)$$
A: your inequality is true, since it is equivalent to
$$3/4\,{\frac { \left( {a}^{2}b+{a}^{2}c+a{b}^{2}-6\,abc+a{c}^{2}+{b}^{2
}c+b{c}^{2} \right) ^{2}}{ \left( -c+a \right) ^{2} \left( b-c
 \right) ^{2} \left( a-b \right) ^{2}}}
\geq 0$$ which is obviously true.
