Already posted right below!

ok, @ssilwa what is answer? I think $ \frac{3}{\sqrt[3]{4}} $.

that's right mathuz

crazyfehmy wrote: Find the maximum value of $M$ for which for all positive real numbers $a, b, c$ we have \[ a^3+b^3+c^3-3abc \geq M(ab^2+bc^2+ca^2-3abc) \] Dear crazyfehmy,Best is this: \[(a^3+b^3+c^3-3abc)^3 \geq \frac{27}{4}(ab^2+bc^2+ca^2-3abc)^3\] But who cna say out conditions? Yes,get "=" conditions ! BQ

See also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=564702

We claim that the answer is $t = \frac{3}{4^{\frac13}}$. We can see that any $M > t$ does not work by letting $a = \epsilon, b = 1, c = 2^{\frac13}$ for some very small $\epsilon.$ Let's now show that $M = \frac{3}{4^{\frac13}}$ actually works. By scaling and cyclic symmetry, we can WLOG assume that $a = 1$ and $a$ is the smallest. Hence, there exist $x, y \in \mathbb{R}_{\ge 0}$ so that $b = 1+x, c = 1+y.$ We now want to show that: $$6+6x+6y+3x^2+3y^2+3xy+x^3+y^3 \ge t(3+3x+3y+x^2+y^2+2xy+xy^2).$$ We will show this by considering the terms of each degree separately. It's clear that $6 \ge 3t$ and $6x+6y \ge 3tx + 3ty.$ It's easy to see by AM-GM that $3x^2+3y^2+3xy \ge tx^2 + ty^2 + 2txy$, because $2(3-t) \ge 2t-3.$ Let's now show that $x^3 + y^3 \ge txy^2.$ By dividing $xy^2$, we want to show that: $$(\frac{x}{y})^2 + \frac{y}{x} \ge t.$$ Letting $z = \frac{x}{y}$, we want to show that $z^2 + \frac{1}{z} \ge t.$ Notice that $z^2 + \frac{1}{z} = z^2 + \frac{1}{2z} + \frac{1}{2z} \ge t$ by AM-GM, and so we're done. $\square$