<s>[INST] 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 .!
 [/INST] 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)
$.
</s>