prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$ prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$ .
I think that must I use from $\dfrac{a^2}{2}+\dfrac{b^2}{2} \geq ab$ but no result please help me .!

I'd like to add a calculus approach to the above answers. If we define the function
$$
f(a,b,c) = \dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} - abc
$$
and look for the positions in $\mathbb{R}_{>0}^3$ where the gradient vanishes, we will find
$$
\overline{\triangledown } f = 0 \Rightarrow a=b=c=1
$$
The value of $f$ at $(1,1,1)$ is 0. However, the function $f$ happens to be "too symmetric" so that the Hessian matrix at $(1,1,1)$ has zero determinant, and thus the second derivative test gives no information about the nature of this critical point (the matrix has a zero eigenvalue).
Nevertheless, we can try a different "version" of the above function, putting $1/c$ in place of $c$. 
$$
g(a,b,c) = \dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{1}{6 c^6} - \dfrac{ab}{c}
$$
Of course, this small transformation keeps (1,1,1) as the only critical point and leaves its nature unchanged. Checking the Hessian for this function at $(1,1,1)$ we get
$$H = \begin{bmatrix}
1 & -1 & 1 \\
-1 & 2 & 1 \\
1 & 1 & 7
\end{bmatrix}$$
which, using Sylvester's Criterion, is positive definite. Thus the point $(1,1,1)$ is a local minimum and since there is no other critical point in the area of interest, it is also a global minimum. That reveals that
$$
f(a,b,c) = \dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} - abc >= 0
$$
for all $(a,b,c) \in \mathbb{R}_{>0}^3$
