gobathegreat wrote: Let $a$,$b$ and $c$ be sides of a triangle such that $a+b+c\le2$. Prove that $-3<{\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}-\frac{a^3}{c}-\frac{b^3}{a}-\frac{c^3}{b}}<3$ Let $a=y+z$, $b=x+z$ and $c=x+y$ where $x\geq y\geq z>0$. Hence, we need to prove that $(x-y)(x-z)(y-z)\sum_{cyc}(3x^2+5xy)<3(x+y)(x+z)(y+z)(x+y+z)^2$. Easy to show that $(x-z)(y-z)\sum_{cyc}(3x^2+5xy)\leq xy(3x^2+3y^2+5xy)$. Id est, it remains to prove that $(x-y)xy(3x^2+3y^2+5xy)<3xy(x+y)^3$, which is obvious.

It is enough to prove that \[ \left|\frac{a^{3}}{b}+\frac{b^{3}}{c}+\frac{c^{3}}{a}-\frac{a^{3}}{c}-\frac{b^{3}}{a}-\frac{c^{3}}{b}\right|<3. \] Notice that \[ \left|\frac{a^{3}}{b}+\frac{b^{3}}{c}+\frac{c^{3}}{a}-\frac{a^{3}}{c}-\frac{b^{3}}{a}-\frac{c^{3}}{b}\right|= \] \[ \left|\frac{(a-b)(b-c)(c-a)\left(a^{2}+ab+ac+b^{2}+bc+c^{2}\right)}{abc}\right|= \] \[ \frac{\left|a-b\right|}{\left|c\right|}\frac{\left|b-c\right|}{\left|a\right|}\frac{\left|c-a\right|}{\left|b\right|}\left(a^{2}+b^{2}+c^{2}+ab+bc+ca\right)< \] \[ a^{2}+b^{2}+c^{2}+ab+bc+ca, \] where in last step we have used triangle inequality. Now, it is enough to prove that \[ a^{2}+b^{2}+c^{2}+ab+bc+ca\leq3 \] or, after homogenization, \[ a^{2}+b^{2}+c^{2}+ab+bc+ca\leq\frac{3}{4}\left(a+b+c\right)^{2}. \] Last inequality is equivalent to well known inequality for triangle sidelengths \[ 2ab+2bc+2ca\geq a^{2}+b^{2}+c^{2} \] that follows after adding up the following obvious inequalities \[ a\left(b+c-a\right)>0, \] \[ b\left(c+a-b\right)>0, \] \[ c\left(a+b-c\right)>0. \]