prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$

$$\frac{a^2}{2}+\frac{b^3}{3}+\frac{c^6}{6}=\frac{a^2}{6}+\frac{a^2}{6}+\frac{a^2}{6}+\frac{b^3}{6}+\frac{b^3}{6}+\frac{c^6}{6}.$$

Then use AM-GM with these $6$ variables.

A slightly different approach using $\frac{a^2}2+\frac{b^2}2\ge ab$ : $$\begin{align*} \frac{a^2}{2}+\frac{b^3}{3}+\frac{c^6}{6}+\frac{abc}3&=\frac13 a^2+\frac13 abc+\left(\frac16 a^2+ \frac16 b^3\right)+\left(\frac16 b^3+\frac16 c^6\right)\\&\ge\frac13a^2+\frac13abc+\frac13ab^{3/2}+\frac13b^{3/2}c^3 \\ &\ge\frac23a^{3/2}b^{1/2}c^{1/2}+\frac23a^{1/2}b^{3/2}c^{3/2}\\ &\ge \frac43abc. \end{align*}$$

This can be solved using the weighted AM-GM inequality. For the case of three variables, this would say: $$x^{\alpha} y^{\beta} z^{\gamma} \le \alpha x + \beta y + \gamma z$$ given that $x,y,z > 0$; $\alpha, \beta, \gamma \ge 0$; and $\alpha + \beta + \gamma = 1$. Now, plug in $x = a^2$, $\alpha = \frac{1}{2}$, $y=b^3$, $\beta = \frac{1}{3}$, $z=c^6$, $\gamma = \frac{1}{6}$.

(Furthermore, if $\alpha, \beta, \gamma$ are all strictly positive, as in this application, then we have that equality holds if and only if $x = y = z$.)