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 .!
 A: A slightly different approach using $\frac{a^2}2+\frac{b^2}2\ge ab$ :
$$\begin{align*}
\frac{a^2}{2}+\frac{b^3}{3}+\frac{c^6}{6}+\frac{abc}3&=\frac13 a^2+\frac13 abc+\left(\frac16 a^2+ \frac16 b^3\right)+\left(\frac16 b^3+\frac16 c^6\right)\\&\ge\frac13a^2+\frac13abc+\frac13ab^{3/2}+\frac13b^{3/2}c^3 \\
&\ge\frac23a^{3/2}b^{1/2}c^{1/2}+\frac23a^{1/2}b^{3/2}c^{3/2}\\
&\ge \frac43abc.
\end{align*}$$
A: 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 $x = y = z$.)
A: 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 $(a,b,c) \in \mathbb{R}_{>0}^3$
A: In general, if
$$\sum_{i=1}^n \frac{1}{p_i} = 1$$
for positive $p_i$, then Jensen's inequality implies that for any concave function $\varphi$ defined on $\mathbb{R}^+$, we have
$$\varphi\left(\sum_{i=1}^n \frac{x_i^{p_i}}{p_i}\right) \ge \sum_{i=1}^n \frac{1}{p_i} \varphi(x_i^{p_i}),$$
for positive $x_i$.  In particular, if $\varphi(x) = \log x$, then we obtain Young's inequality:
$$\sum_{i=1}^n \frac{x_i^{p_i}}{p_i} \ge \prod_{i=1}^n x_i.$$
A: $$\frac{a^2}{2}+\frac{b^3}{3}+\frac{c^6}{6}=\frac{a^2}{6}+\frac{a^2}{6}+\frac{a^2}{6}+\frac{b^3}{6}+\frac{b^3}{6}+\frac{c^6}{6}.$$
Then use AM-GM with these $6$ variables.
A: I naturally want to generalize this.
Here's a first cut.
$\begin{array}\\
\dfrac{a^u}{u}+\dfrac{b^v}{v}
+\dfrac{c^{uv}}{uv}
&=\dfrac{va^u+ub^v+c^{uv}}{uv}\\
&=\dfrac{va^u+ub^v+c^{uv}}{u+v+1}\dfrac{u+v+1}{uv}\\
&\ge (a^{uv}b^{uv}c^{uv})^{1/(u+v+1)}\dfrac{u+v+1}{uv}\\
&= (abc)^{uv/(u+v+1)}\dfrac{u+v+1}{uv}\\
&=\dfrac{(abc)^k}{k}
\qquad k = \dfrac{uv}{u+v+1}\\
\end{array}
$
with equality only if
$a^u = b^v = c^{uv}$
or
$a=c^v$ and $b=c^u$.
From now on
I assume that
$u \le v$.
If $k=1$,
so that
$uv = u+v+1$,
or
$(u-1)(v-1) = 2$,
then
$\dfrac{a^u}{u}+\dfrac{b^v}{v}
+\dfrac{c^{uv}}{uv}
\ge abc
$.
If
$uv = k(u+v+1)$,
or
$(u-k)(v-k) = k^2+k$,
then
$\dfrac{a^u}{u}+\dfrac{b^v}{v}
+\dfrac{c^{uv}}{k}
\ge \dfrac{(abc)^k}{k}
$.
Since
$k^2+k = k(k+1)$,
two solutions to
$(u-k)(v-k) = k^2+k$
are
$u=2k, v=2k+1$
and
$u=k+1, v=k^2+2k$.
To find all the integer solutions,
write
$k^2+k = rs$.
Then
$u=r+k, v=s+k$
are solutions.
The ones above are for
$r=1, s=k^2+k$
and
$r=k, s=k+1$.
These are the same for $k=1$.
For $k=2$,
$rs = 6$
so
$(r, s)
=(1, 6),(2, 3)$
and
$(u, v)
=(3, 8), (4, 5)$.
For
$k=3$,
$rs = 12$
so
$(r, s)
=(1, 12), (2, 6).
(3, 4)
$
and
$(u, v)
=(4, 15), (5, 9),
(6, 7)
$.
